এ অধ্যায়ে আমরা যে বিষয়গুলি আলোচনা করব।
- প্রতিস্থাপন পদ্ধতি ( Method of Replacement )
- কতিপয় স্মরণীয় ফাংশনের যোগজ ( Some memorable Integrals of functions )
- \(\int{(ax+b)^ndx}\)\(=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\)
- \(\int{e^{ax}dx}\)\(=\frac{1}{a}e^{ax}+c\)
- \(\int{\frac{1}{ax+b}dx}\)\(=\frac{1}{a}\ln{|ax+b|}+c\)
- \(\int{\cos{ax}dx}\)\(=\frac{1}{a}\sin{ax}+c\)
- \(\int{\sin{ax}dx}\)\(=-\frac{1}{a}\cos{ax}+c\)
- \(\int{\sec^2{ax}dx}\)\(=\frac{1}{a}\tan{ax}+c\)
- \(\int{e^{ax+b}dx}\)\(=\frac{1}{a}e^{ax+b}+c\)
- \(\int{\frac{1}{(ax+b)^2}dx}\)\(=-\frac{1}{a}\frac{1}{(ax+b)}+c\)
- \(\int{cosec^2{ax}dx}\)\(=-\frac{\cot{ax}}{a}+c\)
- \(\int{\sec{ax}\tan{ax}dx}\)\(=\frac{\sec{ax}}{a}+c\)
- \(\int{cosec \ {ax}\cot{ax}dx}\)\(=-\frac{cosec \ {ax}}{a}+c\)
- \(\int{\cos{(ax+b)}dx}\)\(=\frac{\sin{(ax+b)}}{a}+c\)
- \(\int{\sin{(ax+b)}dx}\)\(=-\frac{\cos{(ax+b)}}{a}+c\)
- \(\int{\sec^2{(ax+b)}dx}\)\(=\frac{1}{a}\tan{(ax+b)}+c\)
- \(\int{a^{mx+n}dx}\)\(=\frac{a^{mx+n}}{m\ln{a}}+c\)
- সমাধানকৃত উদাহরণমালা
- অতি সংক্ষিপ্ত প্রশ্ন-উত্তর
- সংক্ষিপ্ত প্রশ্ন-উত্তর
- বর্ণনামূলক প্রশ্ন-উত্তর
প্রতিস্থাপন পদ্ধতি ( Method of Replacement )
যোগজীকরণ প্রক্রিয়ায় অনেক সময় প্রদত্ত ফাংশনের সরাসরি যোগজ নির্ণয় করা কঠিন হয়ে পড়ে। সেই ক্ষেত্রে প্রতিস্থাপন পদ্ধতি যোগজীকরণ প্রক্রিয়াকে সহজ করে দেয়।
প্রদত্ত যোজ্য রাশি এর অন্তর্ভুক্ত কোনো ফাংশনের পরিবর্তে একটি চলরাশি স্থাপন করাকে প্রতিস্থাপন পদ্ধতি বলে।\(\int{f(ax+b)dx}\) এর ক্ষেত্রে \(ax+b\) কে \(t\) ধরতে হয়।
ধরি,
\(ax+b=t\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(t)\)
\(\Rightarrow a.1+0=\frac{dt}{dx}\)
\(\Rightarrow a=\frac{dt}{dx}\)
\(\Rightarrow adx=dt\)
\(\therefore dx=\frac{1}{a}dt\)
\(\int{f(ax+b)dx}\)\(ax+b=t\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(t)\)
\(\Rightarrow a.1+0=\frac{dt}{dx}\)
\(\Rightarrow a=\frac{dt}{dx}\)
\(\Rightarrow adx=dt\)
\(\therefore dx=\frac{1}{a}dt\)
\(=\int{f(t).\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{f(t)dt}\)
\(\therefore \int{f(ax+b)dx}=\frac{1}{a}\int{f(t)dt}\) এর পর প্রমিত ফাংশনের সূত্র প্রয়োগ করে যোগজীকরণ করতে হয়।
কতিপয় স্মরণীয় ফাংশনের যোগজ ( Some memorable Integrals of functions )
\((1.)\) প্রমাণ কর যে, \(\int{(ax+b)^ndx}=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\)
Proof:
ধরি,
\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{(ax+b)^ndx}\)\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{z^n.\frac{1}{a}dz}\)
\(=\frac{1}{a}.\int{z^ndz}\)
\(=\frac{1}{a}.\frac{z^{n+1}}{n+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
\((2.)\) প্রমাণ কর যে, \(\int{e^{ax}dx}=\frac{1}{a}e^{ax}+c\)
Proof:
ধরি,
\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{e^{ax}dx}\)\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{e^z.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{e^zdz}\)
\(=\frac{1}{a}e^z+c\) ➜ \(\because \int{e^xdx}=e^x+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
\((3.)\) প্রমাণ কর যে, \(\int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}+c \)
Proof:
ধরি,
\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\frac{1}{ax+b}dx}\)\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\frac{1}{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\frac{1}{z}dz}\)
\(=\frac{1}{a}\ln{|z|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|ax+b|}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
\((4.)\) প্রমাণ কর যে, \(\int{\cos{ax}dx}=\frac{1}{a}\sin{ax}+c\)
Proof:
ধরি,
\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\cos{ax}dx}\)\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\cos{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\cos{z}dz}\)
\(=\frac{1}{a}\sin{z}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\sin{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
\((5.)\) প্রমাণ কর যে, \(\int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}+c\)
Proof:
ধরি,
\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\sin{ax}dx}\)\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\sin{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sin{z}dz}\)
\(=\frac{1}{a}(-\cos{z})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{a}\cos{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
\((6.)\) প্রমাণ কর যে, \(\int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}+c\)
Proof:
ধরি,
\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\sec^2{ax}dx}\)\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\sec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec^2{z}dz}\)
\(=\frac{1}{a}\tan{z}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\tan{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
\((7.)\) প্রমাণ কর যে, \(\int{e^{ax+b}dx}=\frac{1}{a}e^{ax+b}+c\)
Proof:
ধরি,
\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{e^{ax+b}dx}\)\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{e^{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{e^{z}dz}\)
\(=\frac{1}{a}e^{z}+c\) ➜ \(\because \int{e^{x}dx}=e^{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{ax+b}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
\((8.)\) প্রমাণ কর যে, \(\int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}\frac{1}{(ax+b)}+c\)
Proof:
ধরি,
\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\frac{1}{(ax+b)^2}dx}\)\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\frac{1}{z^2}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\frac{1}{z^2}dz}\)
\(=\frac{1}{a}\int{z^{-2}dz}\)
\(=\frac{1}{a}\frac{z^{-2+1}}{-2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\frac{z^{-1}}{-1}+c\)
\(=-\frac{1}{a}z^{-1}+c\)
\(=-\frac{1}{a}\frac{1}{z}+c\)
\(=-\frac{1}{a}\frac{1}{ax+b}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
\((9.)\) প্রমাণ কর যে, \(\int{cosec^2{ax}dx}=-\frac{\cot{ax}}{a}+c\)
Proof:
ধরি,
\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{cosec^2{ax}dx}\)\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{cosec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{cosec^2{z}dz}\)
\(=-\frac{1}{a}\cot{z}+c\) ➜ \(\because \int{cosec^2{x}dx}=-\cot{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cot{ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
\((10.)\) প্রমাণ কর যে, \(\int{\sec{ax}\tan{ax}dx}=\frac{\sec{ax}}{a}+c\)
Proof:
ধরি,
\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\sec{ax}\tan{ax}dx}\)\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\sec{z}\tan{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec{z}\tan{z}dz}\)
\(=\frac{1}{a}\sec{z}+c\) ➜ \(\because \int{\sec{x}\tan{x}dx}=\sec{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\sec{ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
\((11.)\) প্রমাণ কর যে, \(\int{cosec \ {ax}\cot{ax}dx}=-\frac{cosec \ {ax}}{a}+c\)
Proof:
ধরি,
\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{cosec \ {ax}\cot{ax}dx}\)\(ax=z\)
\(\Rightarrow \frac{d}{dx}(ax)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{cosec \ {z}\cot{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{cosec \ {z}\cot{z}dz}\)
\(=\frac{1}{a}(-cosec \ {z})+c\) ➜ \(\because \int{cosec \ {x}\cot{x}dx}=-cosec \ {x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{cosec \ {ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
\((12.)\) প্রমাণ কর যে, \(\int{\cos{(ax+b)}dx}=\frac{\sin{(ax+b)}}{a}+c\)
Proof:
ধরি,
\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\cos{(ax+b)}dx}\)\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\cos{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\cos{z}dz}\)
\(=\frac{1}{a}\sin{z}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\sin{(ax+b)}}{a}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
\((13.)\) প্রমাণ কর যে, \(\int{\sin{(ax+b)}dx}=-\frac{\cos{(ax+b)}}{a}+c\)
Proof:
ধরি,
\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\sin{(ax+b)}dx}\)\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\sin{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sin{z}dz}\)
\(=\frac{1}{a}(-\cos{z})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cos{(ax+b)}}{a}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
\((14.)\) প্রমাণ কর যে, \(\int{\sec^2{(ax+b)}dx}=\frac{1}{a}\tan{(ax+b)}+c\)
Proof:
ধরি,
\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(L.H=\int{\sec^2{(ax+b)}dx}\)\(ax+b=z\)
\(\Rightarrow \frac{d}{dx}(ax+b)=\frac{d}{dx}(z)\)
\(\Rightarrow a.1+0=\frac{dz}{dx}\)
\(\Rightarrow a=\frac{dz}{dx}\)
\(\Rightarrow adx=dz\)
\(\therefore dx=\frac{1}{a}dz\)
\(=\int{\sec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec^2{z}dz}\)
\(=\frac{1}{a}\tan{z}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\tan{(ax+b)}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
\((15.)\) প্রমাণ কর যে, \(\int{a^{mx+n}dx}=\frac{a^{mx+n}}{m\ln{a}}+c\)
Proof:
ধরি,
\(mx+n=z\)
\(\Rightarrow \frac{d}{dx}(mx+n)=\frac{d}{dx}(z)\)
\(\Rightarrow m.1+0=\frac{dz}{dx}\)
\(\Rightarrow m=\frac{dz}{dx}\)
\(\Rightarrow mdx=dz\)
\(\therefore dx=\frac{1}{m}dz\)
\(L.H=\int{a^{mx+n}dx}\)\(mx+n=z\)
\(\Rightarrow \frac{d}{dx}(mx+n)=\frac{d}{dx}(z)\)
\(\Rightarrow m.1+0=\frac{dz}{dx}\)
\(\Rightarrow m=\frac{dz}{dx}\)
\(\Rightarrow mdx=dz\)
\(\therefore dx=\frac{1}{m}dz\)
\(=\int{a^{z}.\frac{1}{m}dz}\)
\(=\frac{1}{m}\int{a^{z}dz}\)
\(=\frac{1}{m}\frac{a^{z}}{\ln{a}}+c\) ➜ \(\because \int{a^xdx}=\frac{a^x}{\ln{a}}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{a^{z}}{m\ln{a}}+c\)
\(=\frac{a^{mx+n}}{m\ln{a}}+c\) ➜ \(\because z=mx+n\)
\(=R.H\)
(Proved)
অনুশীলনী \(10.B\) উদাহরণ সমুহ
যোজিত ফল নির্ণয় করঃ
\((1.)\) \(\int{(2x+3)^5dx}\)
উত্তরঃ \(\frac{1}{12}(2x+3)^{6}+c\)
\((2.)\) \(\int{\sin{x^{o}}dx}\)
উত্তরঃ \(-\frac{180}{\pi}\cos{\frac{\pi{x}}{180}}+c\)
\((3.)\) \(\int{\sin{5x}\sin{3x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; চঃ২০১২]
\((4.)\) \(\int{\cos^3{x}dx}\)
উত্তরঃ \(\frac{1}{12}(\sin{3x}+9\sin{x})+c\)
উত্তরঃ \(\frac{1}{12}(2x+3)^{6}+c\)
\((2.)\) \(\int{\sin{x^{o}}dx}\)
উত্তরঃ \(-\frac{180}{\pi}\cos{\frac{\pi{x}}{180}}+c\)
\((3.)\) \(\int{\sin{5x}\sin{3x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; চঃ২০১২]
\((4.)\) \(\int{\cos^3{x}dx}\)
উত্তরঃ \(\frac{1}{12}(\sin{3x}+9\sin{x})+c\)
\((5.)\) \(\int{\cos^4{x}dx}\)
উত্তরঃ \(\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
[ ঢাঃ২০১৪; দিঃ ২০১৩; সিঃ ২০০৮,২০০৪; রাঃ ২০১৪,২০০৭; চঃ ২০০৫ ]
\((6.)\) \(\int{\frac{1}{\sqrt{x}-\sqrt{x-1}}dx}\)
উত্তরঃ \(\frac{2}{3}\left\{x^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right\}+c\)
\((7.)\) \(\int{(1-2x)^4dx}\)
উত্তরঃ \(-\frac{1}{10}(1-2x)^{5}+c\)
উত্তরঃ \(\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
[ ঢাঃ২০১৪; দিঃ ২০১৩; সিঃ ২০০৮,২০০৪; রাঃ ২০১৪,২০০৭; চঃ ২০০৫ ]
\((6.)\) \(\int{\frac{1}{\sqrt{x}-\sqrt{x-1}}dx}\)
উত্তরঃ \(\frac{2}{3}\left\{x^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right\}+c\)
\((7.)\) \(\int{(1-2x)^4dx}\)
উত্তরঃ \(-\frac{1}{10}(1-2x)^{5}+c\)
\((1.)\) \(\int{(2x+3)^5dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(2x+3)^{6}+c\)
উত্তরঃ \(\frac{1}{12}(2x+3)^{6}+c\)
সমাধানঃ
ধরি,
\(2x+3=t\)
\(\Rightarrow \frac{d}{dx}(2x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1+0=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(\int{(2x+3)^5dx}\)\(2x+3=t\)
\(\Rightarrow \frac{d}{dx}(2x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1+0=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(=\int{t^5.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{t^5dt}\)
\(=\frac{1}{2}\frac{t^{5+1}}{5+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{t^{6}}{6}+c\)
\(=\frac{1}{12}t^{6}+c\)
\(=\frac{1}{12}(2x+3)^{6}+c\) ➜ \(\because t=2x+3\)
বিকল্প পদ্ধতিঃ
\(\int{(2x+3)^5dx}\)
\(=\frac{1}{2}\frac{(2x+3)^{5+1}}{5+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(2x+3)^{6}}{6}+c\)
\(=\frac{1}{12}(2x+3)^{6}+c\)
\(=\frac{1}{2}\frac{(2x+3)^{5+1}}{5+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(2x+3)^{6}}{6}+c\)
\(=\frac{1}{12}(2x+3)^{6}+c\)
\((2.)\) \(\int{\sin{x^{o}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{180}{\pi}\cos{\frac{\pi{x}}{180}}+c\)
উত্তরঃ \(-\frac{180}{\pi}\cos{\frac{\pi{x}}{180}}+c\)
সমাধানঃ
আমরা জানি,
\(180^{o}=\pi\)
\(\Rightarrow 1^{o}=\frac{\pi}{180}\)
\(\therefore x^{o}=\frac{\pi{x}}{180}\)
ধরি,
\(\frac{\pi{x}}{180}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{\pi{x}}{180}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{\pi}{180}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{\pi}{180}=\frac{dt}{dx}\)
\(\Rightarrow \frac{\pi}{180}dx=dt\)
\(\therefore dx=\frac{180}{\pi}dt\)
\(\int{\sin{x^{o}}dx}\)\(180^{o}=\pi\)
\(\Rightarrow 1^{o}=\frac{\pi}{180}\)
\(\therefore x^{o}=\frac{\pi{x}}{180}\)
ধরি,
\(\frac{\pi{x}}{180}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{\pi{x}}{180}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{\pi}{180}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{\pi}{180}=\frac{dt}{dx}\)
\(\Rightarrow \frac{\pi}{180}dx=dt\)
\(\therefore dx=\frac{180}{\pi}dt\)
\(=\int{\sin{\frac{\pi{x}}{180}}dx}\)
\(=\int{\sin{t}.\frac{180}{\pi}dt}\)
\(=\frac{180}{\pi}\int{\sin{t}dt}\)
\(=\frac{180}{\pi}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{180}{\pi}\cos{t}+c\)
\(=-\frac{180}{\pi}\cos{\frac{\pi{x}}{180}}+c\) ➜ \(\because t=\frac{\pi{x}}{180}\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{x^{o}}dx}\)\(=\int{\sin{\frac{\pi{x}}{180}}dx}\)
\(=-\frac{1}{\frac{\pi}{180}}\cos{\frac{\pi{x}}{180}}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{180}{\pi}\cos{\frac{\pi{x}}{180}}+c\)
\((3.)\) \(\int{\sin{5x}\sin{3x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; চঃ২০১২]
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; চঃ২০১২]
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(8x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 8.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8dx=dt_{2}\)
\(\therefore dx=\frac{1}{8}dt_{2}\)
\(\int{\sin{5x}\sin{3x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(8x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 8.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8dx=dt_{2}\)
\(\therefore dx=\frac{1}{8}dt_{2}\)
\(=\frac{1}{2}\int{2\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(5x-3x)}-\cos{(5x+3x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\{\cos{(2x)}-\cos{(8x)}\}dx}\)
\(=\frac{1}{2}\int{\cos{(2x)}dx}-\frac{1}{2}\int{\cos{(8x)}dx}\)
\(=\frac{1}{2}\int{\cos{t_{1}}\frac{1}{2}dt_{1}}-\frac{1}{2}\int{\cos{t_{2}}\frac{1}{8}dt_{2}}\)
\(=\frac{1}{4}\int{\cos{t_{1}}dt_{1}}-\frac{1}{16}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{4}\sin{t_{1}}-\frac{1}{16}\sin{t_{2}}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{16}\sin{8x}+c\) ➜ \(\because t_{1}=2x, t_{2}=8x\)
\(=\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{2\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(5x-3x)}-\cos{(5x+3x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\{\cos{(2x)}-\cos{(8x)}\}dx}\)
\(=\frac{1}{2}\int{\cos{(2x)}dx}-\frac{1}{2}\int{\cos{(8x)}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{2}.\frac{1}{8}\sin{8x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{16}\sin{8x}+c\)
\(=\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
\(=\frac{1}{2}\int{2\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(5x-3x)}-\cos{(5x+3x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\{\cos{(2x)}-\cos{(8x)}\}dx}\)
\(=\frac{1}{2}\int{\cos{(2x)}dx}-\frac{1}{2}\int{\cos{(8x)}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{2}.\frac{1}{8}\sin{8x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{16}\sin{8x}+c\)
\(=\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
\((4.)\) \(\int{\cos^3{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(\sin{3x}+9\sin{x})+c\)
উত্তরঃ \(\frac{1}{12}(\sin{3x}+9\sin{x})+c\)
সমাধানঃ
ধরি,
\(3x=t\)
\(\Rightarrow \frac{d}{dx}(3x)=\frac{d}{dx}(t)\)
\(\Rightarrow 3.1=\frac{dt}{dx}\)
\(\Rightarrow 3=\frac{dt}{dx}\)
\(\Rightarrow 3dx=dt\)
\(\therefore dx=\frac{1}{3}dt\)
\(\int{\cos^3{x}dx}\)\(3x=t\)
\(\Rightarrow \frac{d}{dx}(3x)=\frac{d}{dx}(t)\)
\(\Rightarrow 3.1=\frac{dt}{dx}\)
\(\Rightarrow 3=\frac{dt}{dx}\)
\(\Rightarrow 3dx=dt\)
\(\therefore dx=\frac{1}{3}dt\)
\(=\frac{1}{4}\int{4\cos^3{x}dx}\)
\(=\frac{1}{4}\int{(\cos{3x}+3\cos{x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int{\cos{3x}dx}+\frac{3}{4}\int{\cos{x}dx}\)
\(=\frac{1}{4}\int{\cos{t}\frac{1}{3}dt}+\frac{3}{4}\int{\cos{x}dx}\)
\(=\frac{1}{12}\int{\cos{t}dt}+\frac{3}{4}\int{\cos{x}dx}\)
\(=\frac{1}{12}\sin{t}+\frac{3}{4}\sin{x}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{12}\sin{3x}+\frac{3}{4}\sin{x}+c\) ➜ \(\because t=3x\)
\(=\frac{1}{12}(\sin{3x}+9\sin{x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\cos^3{x}dx}\)
\(=\frac{1}{4}\int{4\cos^3{x}dx}\)
\(=\frac{1}{4}\int{(\cos{3x}+3\cos{x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int{\cos{3x}dx}+\frac{3}{4}\int{\cos{x}dx}\)
\(=\frac{1}{4}.\frac{1}{3}\sin{3x}+\frac{3}{4}\sin{x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{12}\sin{3x}+\frac{3}{4}\sin{x}+c\)
\(=\frac{1}{12}(\sin{3x}+9\sin{x})+c\)
\(=\frac{1}{4}\int{4\cos^3{x}dx}\)
\(=\frac{1}{4}\int{(\cos{3x}+3\cos{x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int{\cos{3x}dx}+\frac{3}{4}\int{\cos{x}dx}\)
\(=\frac{1}{4}.\frac{1}{3}\sin{3x}+\frac{3}{4}\sin{x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{12}\sin{3x}+\frac{3}{4}\sin{x}+c\)
\(=\frac{1}{12}(\sin{3x}+9\sin{x})+c\)
\((5.)\) \(\int{\cos^4{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
[ ঢাঃ২০১৪; দিঃ ২০১৩; সিঃ ২০০৮,২০০৪; রাঃ ২০১৪,২০০৭; চঃ ২০০৫ ]
উত্তরঃ \(\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
[ ঢাঃ২০১৪; দিঃ ২০১৩; সিঃ ২০০৮,২০০৪; রাঃ ২০১৪,২০০৭; চঃ ২০০৫ ]
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(\int{\cos^4{x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(=\frac{1}{4}\int{(2\cos^2{x})^2dx}\)
\(=\frac{1}{4}\int{(1+\cos{2x})^2dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{t_{1}}\frac{1}{2}dt_{1}}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{t_{2}}\frac{1}{4}dt_{2}}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{4}\int{\cos{t_{1}}dt_{1}}+\frac{1}{8}\int{dx}+\frac{1}{32}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{4}x+\frac{1}{4}\sin{t_{1}}+\frac{1}{8}x+\frac{1}{32}\sin{t_{2}}+c\) ➜ \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\) ➜ \(\because t_{1}=2x, t_{2}=4x\)
\(=\frac{2+1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\cos^4{x}dx}\)
\(=\frac{1}{4}\int{(2\cos^2{x})^2dx}\)
\(=\frac{1}{4}\int{(1+\cos{2x})^2dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}x+\frac{1}{2}.\frac{1}{2}\sin{2x}+\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{2+1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
\(=\frac{1}{4}\int{(2\cos^2{x})^2dx}\)
\(=\frac{1}{4}\int{(1+\cos{2x})^2dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}x+\frac{1}{2}.\frac{1}{2}\sin{2x}+\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{2+1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
\((6.)\) \(\int{\frac{1}{\sqrt{x}-\sqrt{x-1}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\left\{x^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{2}{3}\left\{x^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right\}+c\)
সমাধানঃ
ধরি,
\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(\int{\frac{dx}{\sqrt{x}-\sqrt{x-1}}}\)\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x}+\sqrt{x-1})\) গুণ করে।
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x})^2-(\sqrt{x-1})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{x-x+1}dx}\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{1}dx}\)
\(=\int{(\sqrt{x}+\sqrt{x-1})dx}\)
\(=\int{\sqrt{x}dx}+\int{\sqrt{x-1}dx}\)
\(=\int{(x)^{\frac{1}{2}}dx}+\int{(x-1)^{\frac{1}{2}}dx}\)
\(=\int{(x)^{\frac{1}{2}}dx}+\int{t^{\frac{1}{2}}dt}\)
\(=\frac{(x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{(x)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{2}{3}(x)^{\frac{3}{2}}+\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=\frac{2}{3}(x)^{\frac{3}{2}}+\frac{2}{3}(x-1)^{\frac{3}{2}}+c\) ➜ \(\because t=x-1\)
\(=\frac{2}{3}\left\{x^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{dx}{\sqrt{x}-\sqrt{x-1}}}\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x}+\sqrt{x-1})\) গুণ করে।
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x})^2-(\sqrt{x-1})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{x-x+1}dx}\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{1}dx}\)
\(=\int{(\sqrt{x}+\sqrt{x-1})dx}\)
\(=\int{\sqrt{x}dx}+\int{\sqrt{x-1}dx}\)
\(=\int{(x)^{\frac{1}{2}}dx}+\int{(x-1)^{\frac{1}{2}}dx}\)
\(=\frac{(x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{1}\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{(x)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+\frac{(x-1)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{2}{3}(x)^{\frac{3}{2}}+\frac{2}{3}(x-1)^{\frac{3}{2}}+c\)
\(=\frac{2}{3}\left\{x^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right\}+c\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x}+\sqrt{x-1})\) গুণ করে।
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x})^2-(\sqrt{x-1})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{x-x+1}dx}\)
\(=\int{\frac{\sqrt{x}+\sqrt{x-1}}{1}dx}\)
\(=\int{(\sqrt{x}+\sqrt{x-1})dx}\)
\(=\int{\sqrt{x}dx}+\int{\sqrt{x-1}dx}\)
\(=\int{(x)^{\frac{1}{2}}dx}+\int{(x-1)^{\frac{1}{2}}dx}\)
\(=\frac{(x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{1}\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{(x)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+\frac{(x-1)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{2}{3}(x)^{\frac{3}{2}}+\frac{2}{3}(x-1)^{\frac{3}{2}}+c\)
\(=\frac{2}{3}\left\{x^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right\}+c\)
\((7.)\) \(\int{(1-2x)^4dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{10}(1-2x)^{5}+c\)
উত্তরঃ \(-\frac{1}{10}(1-2x)^{5}+c\)
সমাধানঃ
ধরি,
\(1-2x=t\)
\(\Rightarrow \frac{d}{dx}(1-2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2.1=\frac{dt}{dx}\)
\(\Rightarrow -2=\frac{dt}{dx}\)
\(\Rightarrow -2dx=dt\)
\(\therefore dx=-\frac{1}{2}dt\)
\(\int{(1-2x)^4dx}\)\(1-2x=t\)
\(\Rightarrow \frac{d}{dx}(1-2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2.1=\frac{dt}{dx}\)
\(\Rightarrow -2=\frac{dt}{dx}\)
\(\Rightarrow -2dx=dt\)
\(\therefore dx=-\frac{1}{2}dt\)
\(=\int{t^4.\left(-\frac{1}{2}\right)dt}\)
\(=-\frac{1}{2}\int{t^4dt}\)
\(=-\frac{1}{2}\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2}\frac{t^{5}}{5}+c\)
\(=-\frac{1}{10}t^{5}+c\)
\(=-\frac{1}{10}(1-2x)^{5}+c\) ➜ \(\because t=1-2x\)
বিকল্প পদ্ধতিঃ
\(\int{(1-2x)^4dx}\)
\(=\frac{1}{-2}\frac{(1-2x)^{4+1}}{4+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2}\frac{(1-2x)^{5}}{5}+c\)
\(=-\frac{1}{10}(1-2x)^{5}+c\)
\(=\frac{1}{-2}\frac{(1-2x)^{4+1}}{4+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2}\frac{(1-2x)^{5}}{5}+c\)
\(=-\frac{1}{10}(1-2x)^{5}+c\)
অনুশীলনী \(10.B / Q.1\)-এর অতি সংক্ষিপ্ত প্রশ্নসমুহ
যোজিত ফল নির্ণয় করঃ
\(Q.1.(i)\) \(\int{(5x+2)^6dx}\)
উত্তরঃ \(\frac{1}{35}(5x+2)^7+c\)
\(Q.1.(ii)\) \(\int{\sqrt{2x+3}dx}\)
উত্তরঃ \(\frac{1}{3}(2x+3)^{\frac{3}{2}}+c\)
\(Q.1.(iii)\) \(\int{\frac{1}{\sqrt[3]{1-4x}}dx}\)
উত্তরঃ \(-\frac{3}{8}\sqrt[3]{(1-4x)^2}+c\)
[ বুয়েটঃ ২০০৫-২০০৬ ]
\(Q.1.(iv)\) \(\int{\frac{1}{\sqrt[3]{1-6x}}dx}\)
উত্তরঃ \(-\frac{1}{4}\sqrt[3]{(1-6x)^{\frac{3}{2}}}+c\)
[ বুয়েটঃ ২০০৫ ]
\(Q.1.(v)\) \(\int{\frac{x}{\sqrt{x+3}}dx}\)
উত্তরঃ \(\frac{2}{3}(x+3)^{\frac{3}{2}}-6\sqrt{(x+3)}+c\)
\(Q.1.(vi)\) \(\int{\frac{x}{\sqrt{1-x}}dx}\)
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{(1-x)}+c\)
[ বুয়েটঃ ২০০৫-২০০৬; ঢাঃ,চঃ ২০১৪ ]
\(Q.1.(vii)\) \(\int{\left\{\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2}\right\}dx}\)
উত্তরঃ \(\frac{1}{a-x}-\frac{1}{a+x}+c\)
\(Q.1.(viii)\) \(\int{\frac{x^2-2x+3}{(x-1)^2}dx}\)
উত্তরঃ \(x-\frac{2}{x-1}+c\)
\(Q.1.(ix)\) \(\int{\frac{xdx}{x-1}}\)
উত্তরঃ \(x+\ln{|x-1|}+c\)
[ সিঃ ২০১৭ ]
উত্তরঃ \(\frac{1}{35}(5x+2)^7+c\)
\(Q.1.(ii)\) \(\int{\sqrt{2x+3}dx}\)
উত্তরঃ \(\frac{1}{3}(2x+3)^{\frac{3}{2}}+c\)
\(Q.1.(iii)\) \(\int{\frac{1}{\sqrt[3]{1-4x}}dx}\)
উত্তরঃ \(-\frac{3}{8}\sqrt[3]{(1-4x)^2}+c\)
[ বুয়েটঃ ২০০৫-২০০৬ ]
\(Q.1.(iv)\) \(\int{\frac{1}{\sqrt[3]{1-6x}}dx}\)
উত্তরঃ \(-\frac{1}{4}\sqrt[3]{(1-6x)^{\frac{3}{2}}}+c\)
[ বুয়েটঃ ২০০৫ ]
\(Q.1.(v)\) \(\int{\frac{x}{\sqrt{x+3}}dx}\)
উত্তরঃ \(\frac{2}{3}(x+3)^{\frac{3}{2}}-6\sqrt{(x+3)}+c\)
\(Q.1.(vi)\) \(\int{\frac{x}{\sqrt{1-x}}dx}\)
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{(1-x)}+c\)
[ বুয়েটঃ ২০০৫-২০০৬; ঢাঃ,চঃ ২০১৪ ]
\(Q.1.(vii)\) \(\int{\left\{\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2}\right\}dx}\)
উত্তরঃ \(\frac{1}{a-x}-\frac{1}{a+x}+c\)
\(Q.1.(viii)\) \(\int{\frac{x^2-2x+3}{(x-1)^2}dx}\)
উত্তরঃ \(x-\frac{2}{x-1}+c\)
\(Q.1.(ix)\) \(\int{\frac{xdx}{x-1}}\)
উত্তরঃ \(x+\ln{|x-1|}+c\)
[ সিঃ ২০১৭ ]
\(Q.1.(x)\) \(\int{\frac{x}{x+3}dx}\)
উত্তরঃ \(x-3\ln{|x+3|}+c\)
\(Q.1.(xi)\) \(\int{\frac{x-2}{x+5}dx}\)
উত্তরঃ \(x-7\ln{|x+5|}+c\)
\(Q.1.(xii)\) \(\int{(5x+2)^3dx}\)
উত্তরঃ \(\frac{1}{20}(5x+2)^4+c\)
\(Q.1.(xiii)\) \(\int{(2-7x)^4dx}\)
উত্তরঃ \(-\frac{1}{35}(2-7x)^5+c\)
\(Q.1.(xiv)\) \(\int{\sqrt{1-x}dx}\)
উত্তরঃ \(-\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(Q.1.(xv)\) \(\int{\frac{dx}{\sqrt{2-3x}}}\)
উত্তরঃ \(-\frac{2}{3}\sqrt{2-3x}+c\)
\(Q.1.(xvi)\) \(\int{\frac{dx}{(1-x)^2}}\)
উত্তরঃ \(\frac{1}{1-x}+c\)
\(Q.1.(xvii)\) \(\int{\frac{2x+1}{2x+3}dx}\)
উত্তরঃ \(x-\ln{|2x+3|}+c\)
\(Q.1.(xviii)\) \(\int{5(3-x)^{-1}dx}\)
উত্তরঃ \(-5\ln{|3-x|}+c\)
\(Q.1.(xix)\) \(\int{\left(\frac{3}{x-1}-\frac{4}{x-2}\right)dx}\)
উত্তরঃ \(\ln{\left\{\frac{(|x-1|)^3}{(x-2)^4}\right\}}+c\)
উত্তরঃ \(x-3\ln{|x+3|}+c\)
\(Q.1.(xi)\) \(\int{\frac{x-2}{x+5}dx}\)
উত্তরঃ \(x-7\ln{|x+5|}+c\)
\(Q.1.(xii)\) \(\int{(5x+2)^3dx}\)
উত্তরঃ \(\frac{1}{20}(5x+2)^4+c\)
\(Q.1.(xiii)\) \(\int{(2-7x)^4dx}\)
উত্তরঃ \(-\frac{1}{35}(2-7x)^5+c\)
\(Q.1.(xiv)\) \(\int{\sqrt{1-x}dx}\)
উত্তরঃ \(-\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(Q.1.(xv)\) \(\int{\frac{dx}{\sqrt{2-3x}}}\)
উত্তরঃ \(-\frac{2}{3}\sqrt{2-3x}+c\)
\(Q.1.(xvi)\) \(\int{\frac{dx}{(1-x)^2}}\)
উত্তরঃ \(\frac{1}{1-x}+c\)
\(Q.1.(xvii)\) \(\int{\frac{2x+1}{2x+3}dx}\)
উত্তরঃ \(x-\ln{|2x+3|}+c\)
\(Q.1.(xviii)\) \(\int{5(3-x)^{-1}dx}\)
উত্তরঃ \(-5\ln{|3-x|}+c\)
\(Q.1.(xix)\) \(\int{\left(\frac{3}{x-1}-\frac{4}{x-2}\right)dx}\)
উত্তরঃ \(\ln{\left\{\frac{(|x-1|)^3}{(x-2)^4}\right\}}+c\)
\(Q.1.(i)\) \(\int{(5x+2)^6dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{35}(5x+2)^7+c\)
উত্তরঃ \(\frac{1}{35}(5x+2)^7+c\)
সমাধানঃ
ধরি,
\(5x+2=t\)
\(\Rightarrow \frac{d}{dx}(5x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1+0=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(\int{(5x+2)^6dx}\)\(5x+2=t\)
\(\Rightarrow \frac{d}{dx}(5x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1+0=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(=\int{t^6.\frac{1}{5}dt}\)
\(=\frac{1}{5}\int{t^6dt}\)
\(=\frac{1}{5}\frac{t^{6+1}}{6+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\frac{t^{7}}{7}+c\)
\(=\frac{1}{35}t^{7}+c\)
\(=\frac{1}{35}(5x+2)^{7}+c\) ➜\(\because t=5x+2\)
বিকল্প পদ্ধতিঃ
\(\int{(5x+2)^6dx}\)
\(=\frac{1}{5}\frac{(5x+2)^{6+1}}{6+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\frac{(5x+2)^{7}}{7}+c\)
\(=\frac{1}{35}(5x+2)^{7}+c\)
\(=\frac{1}{5}\frac{(5x+2)^{6+1}}{6+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\frac{(5x+2)^{7}}{7}+c\)
\(=\frac{1}{35}(5x+2)^{7}+c\)
\(Q.1.(ii)\) \(\int{\sqrt{2x+3}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}(2x+3)^{\frac{3}{2}}+c\)
উত্তরঃ \(\frac{1}{3}(2x+3)^{\frac{3}{2}}+c\)
সমাধানঃ
ধরি,
\(2x+3=t\)
\(\Rightarrow \frac{d}{dx}(2x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1+0=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(\int{\sqrt{2x+3}dx}\)\(2x+3=t\)
\(\Rightarrow \frac{d}{dx}(2x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1+0=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(=\int{(2x+3)^{\frac{1}{2}}dx}\)
\(=\int{t^{\frac{1}{2}}.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{t^{\frac{1}{2}}dt}\)
\(=\frac{1}{2}\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\)
\(=\frac{1}{2}\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{2}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=\frac{1}{3}t^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(2x+3)^{\frac{3}{2}}+c\) ➜\(\because t=2x+3\)
বিকল্প পদ্ধতিঃ
\(\int{\sqrt{2x+3}dx}\)
\(=\int{(2x+3)^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\frac{(2x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(2x+3)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{2}\frac{(2x+3)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}(2x+3)^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(2x+3)^{\frac{3}{2}}+c\)
\(=\int{(2x+3)^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\frac{(2x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(2x+3)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{2}\frac{(2x+3)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}(2x+3)^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(2x+3)^{\frac{3}{2}}+c\)
\(Q.1.(iii)\) \(\int{\frac{1}{\sqrt[3]{1-4x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{3}{8}\sqrt[3]{(1-4x)^2}+c\)
[ বুয়েটঃ ২০০৫-২০০৬ ]
উত্তরঃ \(-\frac{3}{8}\sqrt[3]{(1-4x)^2}+c\)
[ বুয়েটঃ ২০০৫-২০০৬ ]
সমাধানঃ
ধরি,
\(1-4x=t\)
\(\Rightarrow \frac{d}{dx}(1-4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-4.1=\frac{dt}{dx}\)
\(\Rightarrow -4=\frac{dt}{dx}\)
\(\Rightarrow -4dx=dt\)
\(\therefore dx=-\frac{1}{4}dt\)
\(\int{\frac{1}{\sqrt[3]{1-4x}}dx}\)\(1-4x=t\)
\(\Rightarrow \frac{d}{dx}(1-4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-4.1=\frac{dt}{dx}\)
\(\Rightarrow -4=\frac{dt}{dx}\)
\(\Rightarrow -4dx=dt\)
\(\therefore dx=-\frac{1}{4}dt\)
\(=\int{\frac{1}{(1-4x)^{\frac{1}{3}}}dx}\)
\(=\int{(1-4x)^{-\frac{1}{3}}dx}\)
\(=\int{t^{-\frac{1}{3}}\times{-\frac{1}{4}dt}}\)
\(=-\frac{1}{4}\int{t^{-\frac{1}{3}}dt}\)
\(=-\frac{1}{4}\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\)
\(=-\frac{1}{4}\frac{t^{\frac{-1+3}{3}}}{\frac{-1+3}{3}}+c\)
\(=-\frac{1}{4}\frac{t^{\frac{2}{3}}}{\frac{2}{3}}+c\)
\(=-\frac{1}{4}.\frac{3}{2}t^{\frac{2}{3}}+c\)
\(=-\frac{3}{8}t^{\frac{2}{3}}+c\)
\(=-\frac{3}{8}(1-4x)^{\frac{2}{3}}+c\) ➜\(\because t=1-4x\)
\(=-\frac{3}{8}\sqrt[3]{(1-4x)^2}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1}{\sqrt[3]{1-4x}}dx}\)
\(=\int{\frac{1}{(1-4x)^{\frac{1}{3}}}dx}\)
\(=\int{(1-4x)^{-\frac{1}{3}}dx}\)
\(=\frac{1}{-4}\frac{(1-4x)^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}\frac{(1-4x)^{\frac{-1+3}{3}}}{\frac{-1+3}{3}}+c\)
\(=-\frac{1}{4}\frac{(1-4x)^{\frac{2}{3}}}{\frac{2}{3}}+c\)
\(=-\frac{1}{4}.\frac{3}{2}(1-6x)^{\frac{2}{3}}+c\)
\(=-\frac{3}{8}(1-4x)^{\frac{2}{3}}+c\)
\(=-\frac{3}{8}\sqrt[3]{(1-4x)^2}+c\)
\(=\int{\frac{1}{(1-4x)^{\frac{1}{3}}}dx}\)
\(=\int{(1-4x)^{-\frac{1}{3}}dx}\)
\(=\frac{1}{-4}\frac{(1-4x)^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}\frac{(1-4x)^{\frac{-1+3}{3}}}{\frac{-1+3}{3}}+c\)
\(=-\frac{1}{4}\frac{(1-4x)^{\frac{2}{3}}}{\frac{2}{3}}+c\)
\(=-\frac{1}{4}.\frac{3}{2}(1-6x)^{\frac{2}{3}}+c\)
\(=-\frac{3}{8}(1-4x)^{\frac{2}{3}}+c\)
\(=-\frac{3}{8}\sqrt[3]{(1-4x)^2}+c\)
\(Q.1.(iv)\) \(\int{\frac{1}{\sqrt[3]{1-6x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{4}\sqrt[3]{(1-6x)^{\frac{3}{2}}}+c\)
[ বুয়েটঃ ২০০৫ ]
উত্তরঃ \(-\frac{1}{4}\sqrt[3]{(1-6x)^{\frac{3}{2}}}+c\)
[ বুয়েটঃ ২০০৫ ]
সমাধানঃ
ধরি,
\(1-6x=t\)
\(\Rightarrow \frac{d}{dx}(1-6x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-6.1=\frac{dt}{dx}\)
\(\Rightarrow -6=\frac{dt}{dx}\)
\(\Rightarrow -6dx=dt\)
\(\therefore dx=-\frac{1}{6}dt\)
\(\int{\frac{1}{\sqrt[3]{1-6x}}dx}\)\(1-6x=t\)
\(\Rightarrow \frac{d}{dx}(1-6x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-6.1=\frac{dt}{dx}\)
\(\Rightarrow -6=\frac{dt}{dx}\)
\(\Rightarrow -6dx=dt\)
\(\therefore dx=-\frac{1}{6}dt\)
\(=\int{\frac{1}{(1-6x)^{\frac{1}{3}}}dx}\)
\(=\int{(1-6x)^{-\frac{1}{3}}dx}\)
\(=\int{t^{-\frac{1}{3}}\times{-\frac{1}{6}dt}}\)
\(=-\frac{1}{6}\int{t^{-\frac{1}{3}}dt}\)
\(=-\frac{1}{6}\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{6}\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\)
\(=-\frac{1}{6}\frac{t^{\frac{-1+3}{3}}}{\frac{-1+3}{3}}+c\)
\(=-\frac{1}{6}\frac{t^{\frac{2}{3}}}{\frac{2}{3}}+c\)
\(=-\frac{1}{6}.\frac{3}{2}t^{\frac{2}{3}}+c\)
\(=-\frac{1}{4}t^{\frac{2}{3}}+c\)
\(=-\frac{1}{4}(1-6x)^{\frac{2}{3}}+c\) ➜\(\because t=1-6x\)
\(=-\frac{1}{4}\sqrt[3]{(1-6x)^2}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1}{\sqrt[3]{1-6x}}dx}\)
\(=\int{\frac{1}{(1-6x)^{\frac{1}{3}}}dx}\)
\(=\int{(1-6x)^{-\frac{1}{3}}dx}\)
\(=\frac{1}{-6}\frac{(1-6x)^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{6}\frac{(1-6x)^{\frac{-1+3}{3}}}{\frac{-1+3}{3}}+c\)
\(=-\frac{1}{6}\frac{(1-6x)^{\frac{2}{3}}}{\frac{2}{3}}+c\)
\(=-\frac{1}{6}.\frac{3}{2}(1-6x)^{\frac{2}{3}}+c\)
\(=-\frac{1}{4}(1-6x)^{\frac{2}{3}}+c\)
\(=-\frac{1}{4}\sqrt[3]{(1-6x)^2}+c\)
\(=\int{\frac{1}{(1-6x)^{\frac{1}{3}}}dx}\)
\(=\int{(1-6x)^{-\frac{1}{3}}dx}\)
\(=\frac{1}{-6}\frac{(1-6x)^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{6}\frac{(1-6x)^{\frac{-1+3}{3}}}{\frac{-1+3}{3}}+c\)
\(=-\frac{1}{6}\frac{(1-6x)^{\frac{2}{3}}}{\frac{2}{3}}+c\)
\(=-\frac{1}{6}.\frac{3}{2}(1-6x)^{\frac{2}{3}}+c\)
\(=-\frac{1}{4}(1-6x)^{\frac{2}{3}}+c\)
\(=-\frac{1}{4}\sqrt[3]{(1-6x)^2}+c\)
\(Q.1.(v)\) \(\int{\frac{x}{\sqrt{x+3}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}(x+3)^{\frac{3}{2}}-6\sqrt{(x+3)}+c\)
উত্তরঃ \(\frac{2}{3}(x+3)^{\frac{3}{2}}-6\sqrt{(x+3)}+c\)
সমাধানঃ
ধরি,
\(x+3=t\)
\(\Rightarrow \frac{d}{dx}(x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(\int{\frac{x}{\sqrt{x+3}}dx}\)\(x+3=t\)
\(\Rightarrow \frac{d}{dx}(x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(=\int{\frac{x+3-3}{\sqrt{x+3}}dx}\)
\(=\int{\left(\frac{x+3}{\sqrt{x+3}}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left(\frac{\sqrt{x+3}.\sqrt{x+3}}{\sqrt{x+3}}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left(\sqrt{x+3}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left((x+3)^{\frac{1}{2}}-3\frac{1}{(x+3)^{\frac{1}{2}}}\right)dx}\)
\(=\int{\left((x+3)^{\frac{1}{2}}-3(x+3)^{-\frac{1}{2}}\right)dx}\)
\(=\int{(x+3)^{\frac{1}{2}}dx}-3\int{(x+3)^{-\frac{1}{2}}dx}\)
\(=\int{t^{\frac{1}{2}}dt}-3\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-3\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}-3\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-3\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}-6t^{\frac{1}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}-6\sqrt{t}+c\)
\(=\frac{2}{3}(x+3)^{\frac{3}{2}}-6\sqrt{x+3}+c\) ➜\(\because t=x+3\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{x}{\sqrt{x+3}}dx}\)
\(=\int{\frac{x+3-3}{\sqrt{x+3}}dx}\)
\(=\int{\left(\frac{x+3}{\sqrt{x+3}}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left(\frac{\sqrt{x+3}.\sqrt{x+3}}{\sqrt{x+3}}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left(\sqrt{x+3}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left((x+3)^{\frac{1}{2}}-3\frac{1}{(x+3)^{\frac{1}{2}}}\right)dx}\)
\(=\int{\left((x+3)^{\frac{1}{2}}-3(x+3)^{-\frac{1}{2}}\right)dx}\)
\(=\int{(x+3)^{\frac{1}{2}}dx}-3\int{(x+3)^{-\frac{1}{2}}dx}\)
\(=\frac{1}{1}\frac{(x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-3\frac{1}{1}\frac{(x+3)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{(x+3)^{\frac{1+2}{2}}}{\frac{1+2}{2}}-3\frac{(x+3)^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}-3\frac{(x+3)^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=\frac{2}{3}(x+3)^{\frac{3}{2}}-6(x+3)^{\frac{1}{2}}+c\)
\(=\frac{2}{3}(x+3)^{\frac{3}{2}}-6\sqrt{x+3}+c\)
\(=\int{\frac{x+3-3}{\sqrt{x+3}}dx}\)
\(=\int{\left(\frac{x+3}{\sqrt{x+3}}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left(\frac{\sqrt{x+3}.\sqrt{x+3}}{\sqrt{x+3}}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left(\sqrt{x+3}-\frac{3}{\sqrt{x+3}}\right)dx}\)
\(=\int{\left((x+3)^{\frac{1}{2}}-3\frac{1}{(x+3)^{\frac{1}{2}}}\right)dx}\)
\(=\int{\left((x+3)^{\frac{1}{2}}-3(x+3)^{-\frac{1}{2}}\right)dx}\)
\(=\int{(x+3)^{\frac{1}{2}}dx}-3\int{(x+3)^{-\frac{1}{2}}dx}\)
\(=\frac{1}{1}\frac{(x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-3\frac{1}{1}\frac{(x+3)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{(x+3)^{\frac{1+2}{2}}}{\frac{1+2}{2}}-3\frac{(x+3)^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}-3\frac{(x+3)^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=\frac{2}{3}(x+3)^{\frac{3}{2}}-6(x+3)^{\frac{1}{2}}+c\)
\(=\frac{2}{3}(x+3)^{\frac{3}{2}}-6\sqrt{x+3}+c\)
\(Q.1.(vi)\) \(\int{\frac{x}{\sqrt{1-x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{(1-x)}+c\)
[ বুয়েটঃ ২০০৫-২০০৬; ঢাঃ,চঃ ২০১৪ ]
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{(1-x)}+c\)
[ বুয়েটঃ ২০০৫-২০০৬; ঢাঃ,চঃ ২০১৪ ]
সমাধানঃ
ধরি,
\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(\int{\frac{x}{\sqrt{1-x}}dx}\)\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(=\int{\frac{1-(1-x)}{\sqrt{1-x}}dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\frac{(1-x)}{\sqrt{1-x}}\right)dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\frac{\sqrt{1-x}.\sqrt{1-x}}{\sqrt{1-x}}\right)dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\sqrt{1-x}\right)dx}\)
\(=\int{\left(\frac{1}{(1-x)^{\frac{1}{2}}}-(1-x)^{\frac{1}{2}}\right)dx}\)
\(=\int{\left((1-x)^{-\frac{1}{2}}-(1-x)^{\frac{1}{2}}\right)dx}\)
\(=\int{(1-x)^{-\frac{1}{2}}dx}-\int{(1-x)^{\frac{1}{2}}dx}\)
\(=\int{t^{-\frac{1}{2}}(-dt)}-\int{t^{\frac{1}{2}}(-dt)}\)
\(=-\int{t^{-\frac{1}{2}}dt}+\int{t^{\frac{1}{2}}dt}\)
\(=-\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\)
\(=-\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-2t^{\frac{1}{2}}+\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}-2t^{\frac{1}{2}}+c\)
\(=\frac{2}{3}(1-x)^{\frac{3}{2}}-2(1-x)^{\frac{1}{2}}+c\) ➜\(\because t=1-x\)
\(=\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{1-x}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{x}{\sqrt{1-x}}dx}\)
\(=\int{\frac{1-(1-x)}{\sqrt{1-x}}dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\frac{(1-x)}{\sqrt{1-x}}\right)dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\frac{\sqrt{1-x}.\sqrt{1-x}}{\sqrt{1-x}}\right)dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\sqrt{1-x}\right)dx}\)
\(=\int{\left(\frac{1}{(1-x)^{\frac{1}{2}}}-(1-x)^{\frac{1}{2}}\right)dx}\)
\(=\int{\left((1-x)^{-\frac{1}{2}}-(1-x)^{\frac{1}{2}}\right)dx}\)
\(=\int{(1-x)^{-\frac{1}{2}}dx}-\int{(1-x)^{\frac{1}{2}}dx}\)
\(=\frac{1}{-1}\frac{(1-x)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-\frac{1}{-1}\frac{(1-x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(1-x)^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+\frac{(1-x)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{(1-x)^{\frac{1}{2}}}{\frac{1}{2}}+\frac{(1-x)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-2(1-x)^{\frac{1}{2}}+\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(=-2\sqrt{(1-x)}+\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(=\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{(1-x)}+c\)
\(=\int{\frac{1-(1-x)}{\sqrt{1-x}}dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\frac{(1-x)}{\sqrt{1-x}}\right)dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\frac{\sqrt{1-x}.\sqrt{1-x}}{\sqrt{1-x}}\right)dx}\)
\(=\int{\left(\frac{1}{\sqrt{1-x}}-\sqrt{1-x}\right)dx}\)
\(=\int{\left(\frac{1}{(1-x)^{\frac{1}{2}}}-(1-x)^{\frac{1}{2}}\right)dx}\)
\(=\int{\left((1-x)^{-\frac{1}{2}}-(1-x)^{\frac{1}{2}}\right)dx}\)
\(=\int{(1-x)^{-\frac{1}{2}}dx}-\int{(1-x)^{\frac{1}{2}}dx}\)
\(=\frac{1}{-1}\frac{(1-x)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-\frac{1}{-1}\frac{(1-x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(1-x)^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+\frac{(1-x)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{(1-x)^{\frac{1}{2}}}{\frac{1}{2}}+\frac{(1-x)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-2(1-x)^{\frac{1}{2}}+\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(=-2\sqrt{(1-x)}+\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(=\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{(1-x)}+c\)
\(Q.1.(vii)\) \(\int{\left\{\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2}\right\}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{a-x}-\frac{1}{a+x}+c\)
উত্তরঃ \(\frac{1}{a-x}-\frac{1}{a+x}+c\)
সমাধানঃ
ধরি,
\(a+x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(a+x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 0+1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\therefore dx=dt_{1}\)
আবার,
\(a-x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(a-x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 0-1=\frac{dt_{2}}{dx}\)
\(\Rightarrow -1=\frac{dt_{2}}{dx}\)
\(\Rightarrow -dx=dt_{2}\)
\(\therefore dx=-dt_{2}\)
\(\int{\left\{\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2}\right\}dx}\)\(a+x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(a+x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 0+1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\therefore dx=dt_{1}\)
আবার,
\(a-x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(a-x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 0-1=\frac{dt_{2}}{dx}\)
\(\Rightarrow -1=\frac{dt_{2}}{dx}\)
\(\Rightarrow -dx=dt_{2}\)
\(\therefore dx=-dt_{2}\)
\(=\int{\left\{(a+x)^{-2}+(a-x)^{-2}\right\}dx}\)
\(=\int{(a+x)^{-2}dx}+\int{(a-x)^{-2}dx}\)
\(=\int{t^{-2}_{1}dt_{1}}+\int{t^{-2}_{2}(-dt_{2})}\)
\(=\int{t^{-2}_{1}dt_{1}}-\int{t^{-2}_{2}dt_{2}}\)
\(=\frac{t^{-2+1}_{1}}{-2+1}-\frac{t^{-2+1}_{2}}{-2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{-1}_{1}}{-1}-\frac{t^{-1}_{2}}{-1}+c\)
\(=-t^{-1}_{1}+t^{-1}_{2}+c\)
\(=t^{-1}_{2}-t^{-1}_{1}+c\)
\(=\frac{1}{t_{2}}-\frac{1}{t_{1}}+c\)
\(=\frac{1}{a-x}-\frac{1}{a+x}+c\) ➜\(\because t_{2}=a-x, t_{1}=a+x\)
বিকল্প পদ্ধতিঃ
\(\int{\left\{\frac{1}{(a+x)^2}+\frac{1}{(a-x)^2}\right\}dx}\)
\(=\int{\left\{(a+x)^{-2}+(a-x)^{-2}\right\}dx}\)
\(=\int{(a+x)^{-2}dx}+\int{(a-x)^{-2}dx}\)
\(=\left(-\frac{1}{1}\frac{1}{a+x}\right)+\left(-\frac{1}{-1}\frac{1}{a-x}\right)+c\) ➜ \(\because \int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}\frac{1}{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{a+x}+\frac{1}{a-x}+c\)
\(=\frac{1}{a-x}-\frac{1}{a+x}+c\)
\(=\int{\left\{(a+x)^{-2}+(a-x)^{-2}\right\}dx}\)
\(=\int{(a+x)^{-2}dx}+\int{(a-x)^{-2}dx}\)
\(=\left(-\frac{1}{1}\frac{1}{a+x}\right)+\left(-\frac{1}{-1}\frac{1}{a-x}\right)+c\) ➜ \(\because \int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}\frac{1}{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{a+x}+\frac{1}{a-x}+c\)
\(=\frac{1}{a-x}-\frac{1}{a+x}+c\)
\(Q.1.(viii)\) \(\int{\frac{x^2-2x+3}{(x-1)^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(x-\frac{2}{x-1}+c\)
উত্তরঃ \(x-\frac{2}{x-1}+c\)
সমাধানঃ
ধরি,
\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(\int{\frac{x^2-2x+3}{(x-1)^2}dx}\)\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(=\int{\frac{x^2-2x+1+2}{(x-1)^2}dx}\)
\(=\int{\frac{(x-1)^2+2}{(x-1)^2}dx}\)
\(=\int{\left\{\frac{(x-1)^2}{(x-1)^2}+\frac{2}{(x-1)^2}\right\}dx}\)
\(=\int{\left\{1+2\frac{1}{(x-1)^2}\right\}dx}\)
\(=\int{\left\{1+2(x-1)^{-2}\right\}dx}\)
\(=\int{dx}+2\int{(x-1)^{-2}dx}\)
\(=\int{dx}+2\int{t^{-2}dt}\)
\(=x+2\frac{t^{-2+1}}{-2+1}+c\) ➜ \(\because \int{dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+2\frac{t^{-1}}{-1}+c\)
\(=x-2t^{-1}+c\)
\(=x-2\frac{1}{t}+c\)
\(=x-\frac{2}{x-1}+c\) ➜\(\because t=x-1\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{x^2-2x+3}{(x-1)^2}dx}\)
\(=\int{\frac{x^2-2x+1+2}{(x-1)^2}dx}\)
\(=\int{\frac{(x-1)^2+2}{(x-1)^2}dx}\)
\(=\int{\left\{\frac{(x-1)^2}{(x-1)^2}+\frac{2}{(x-1)^2}\right\}dx}\)
\(=\int{\left\{1+2\frac{1}{(x-1)^2}\right\}dx}\)
\(=\int{dx}+2\int{\frac{1}{(x-1)^2}dx}\)
\(=x+2\times{-\frac{1}{1}\frac{1}{x-1}}+c\) ➜ \(\because \int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}\frac{1}{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-2\frac{1}{x-1}+c\)
\(=x-\frac{2}{x-1}+c\)
\(=\int{\frac{x^2-2x+1+2}{(x-1)^2}dx}\)
\(=\int{\frac{(x-1)^2+2}{(x-1)^2}dx}\)
\(=\int{\left\{\frac{(x-1)^2}{(x-1)^2}+\frac{2}{(x-1)^2}\right\}dx}\)
\(=\int{\left\{1+2\frac{1}{(x-1)^2}\right\}dx}\)
\(=\int{dx}+2\int{\frac{1}{(x-1)^2}dx}\)
\(=x+2\times{-\frac{1}{1}\frac{1}{x-1}}+c\) ➜ \(\because \int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}\frac{1}{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-2\frac{1}{x-1}+c\)
\(=x-\frac{2}{x-1}+c\)
\(Q.1.(ix)\) \(\int{\frac{xdx}{x-1}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(x+\ln{|x-1|}+c\)
[ সিঃ ২০১৭ ]
উত্তরঃ \(x+\ln{|x-1|}+c\)
[ সিঃ ২০১৭ ]
সমাধানঃ
ধরি,
\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(\int{\frac{xdx}{x-1}}\)\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(=\int{\frac{x-1+1}{x-1}dx}\)
\(=\int{\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right)dx}\)
\(=\int{\left(1+\frac{1}{x-1}\right)dx}\)
\(=\int{dx}+\int{\frac{1}{x-1}dx}\)
\(=\int{dx}+\int{\frac{1}{t}dt}\)
\(=x+\ln{|t|}+c\) ➜ \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\ln{|x-1|}+c\) ➜\(\because t=x-1\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{xdx}{x-1}}\)
\(=\int{\frac{x-1+1}{x-1}dx}\)
\(=\int{\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right)dx}\)
\(=\int{\left(1+\frac{1}{x-1}\right)dx}\)
\(=\int{dx}+\int{\frac{1}{x-1}dx}\)
\(=x+\frac{1}{1}\ln{|x-1|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\ln{|x-1|}+c\)
\(=\int{\frac{x-1+1}{x-1}dx}\)
\(=\int{\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right)dx}\)
\(=\int{\left(1+\frac{1}{x-1}\right)dx}\)
\(=\int{dx}+\int{\frac{1}{x-1}dx}\)
\(=x+\frac{1}{1}\ln{|x-1|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\ln{|x-1|}+c\)
\(Q.1.(x)\) \(\int{\frac{x}{x+3}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(x-3\ln{|x+3|}+c\)
উত্তরঃ \(x-3\ln{|x+3|}+c\)
সমাধানঃ
ধরি,
\(x+3=t\)
\(\Rightarrow \frac{d}{dx}(x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(\int{\frac{x}{x+3}dx}\)\(x+3=t\)
\(\Rightarrow \frac{d}{dx}(x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(=\int{\frac{x+3-3}{x+3}dx}\)
\(=\int{\left(\frac{x+3}{x+3}-3\frac{1}{x+3}\right)dx}\)
\(=\int{\left(1-3\frac{1}{x+3}\right)dx}\)
\(=\int{dx}-3\int{\frac{1}{x+3}dx}\)
\(=\int{dx}-3\int{\frac{1}{t}dt}\)
\(=x-3\ln{|t|}+c\) ➜ \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-3\ln{|x+3|}+c\) ➜\(\because t=x+3\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{x}{x+3}dx}\)
\(=\int{\frac{x+3-3}{x+3}dx}\)
\(=\int{\left(\frac{x+3}{x+3}-3\frac{1}{x+3}\right)dx}\)
\(=\int{\left(1-3\frac{1}{x+3}\right)dx}\)
\(=\int{dx}-3\int{\frac{1}{x+3}dx}\)
\(=x-3\frac{1}{1}\ln{|x+3|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-3\ln{|x+3|}+c\)
\(=\int{\frac{x+3-3}{x+3}dx}\)
\(=\int{\left(\frac{x+3}{x+3}-3\frac{1}{x+3}\right)dx}\)
\(=\int{\left(1-3\frac{1}{x+3}\right)dx}\)
\(=\int{dx}-3\int{\frac{1}{x+3}dx}\)
\(=x-3\frac{1}{1}\ln{|x+3|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-3\ln{|x+3|}+c\)
\(Q.1.(xi)\) \(\int{\frac{x-2}{x+5}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(x-7\ln{|x+5|}+c\)
উত্তরঃ \(x-7\ln{|x+5|}+c\)
সমাধানঃ
ধরি,
\(x+5=t\)
\(\Rightarrow \frac{d}{dx}(x+5)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(\int{\frac{x-2}{x+5}dx}\)\(x+5=t\)
\(\Rightarrow \frac{d}{dx}(x+5)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
\(=\int{\frac{x+5-7}{x+5}dx}\)
\(=\int{\left(\frac{x+5}{x+5}-7\frac{1}{x+5}\right)dx}\)
\(=\int{\left(1-7\frac{1}{x+5}\right)dx}\)
\(=\int{dx}-7\int{\frac{1}{x+5}dx}\)
\(=\int{dx}-7\int{\frac{1}{t}dt}\)
\(=x-7\ln{|t|}+c\) ➜ \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-7\ln{|x+5|}+c\) ➜\(\because t=x+5\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{x-2}{x+5}dx}\)
\(=\int{\frac{x+5-7}{x+5}dx}\)
\(=\int{\left(\frac{x+5}{x+5}-7\frac{1}{x+5}\right)dx}\)
\(=\int{\left(1-7\frac{1}{x+5}\right)dx}\)
\(=\int{dx}-7\int{\frac{1}{x+5}dx}\)
\(=x-7\frac{1}{1}\ln{|x+5|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-7\ln{|x+5|}+c\)
\(=\int{\frac{x+5-7}{x+5}dx}\)
\(=\int{\left(\frac{x+5}{x+5}-7\frac{1}{x+5}\right)dx}\)
\(=\int{\left(1-7\frac{1}{x+5}\right)dx}\)
\(=\int{dx}-7\int{\frac{1}{x+5}dx}\)
\(=x-7\frac{1}{1}\ln{|x+5|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-7\ln{|x+5|}+c\)
\(Q.1.(xii)\) \(\int{(5x+2)^3dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{20}(5x+2)^4+c\)
উত্তরঃ \(\frac{1}{20}(5x+2)^4+c\)
সমাধানঃ
ধরি,
\(5x+2=t\)
\(\Rightarrow \frac{d}{dx}(5x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1+0=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(\int{(5x+2)^3dx}\)\(5x+2=t\)
\(\Rightarrow \frac{d}{dx}(5x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1+0=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(=\int{t^3.\frac{1}{5}dt}\)
\(=\frac{1}{5}\int{t^3dt}\)
\(=\frac{1}{5}\frac{t^{3+1}}{3+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\frac{t^{4}}{4}+c\)
\(=\frac{1}{20}t^{4}+c\)
\(=\frac{1}{20}(5x+2)^{4}+c\) ➜\(\because t=5x+2\)
বিকল্প পদ্ধতিঃ
\(\int{(5x+2)^3dx}\)
\(=\frac{1}{5}\frac{(5x+2)^{3+1}}{3+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\frac{(5x+2)^{4}}{4}+c\)
\(=\frac{1}{20}(5x+2)^{4}+c\)
\(=\frac{1}{5}\frac{(5x+2)^{3+1}}{3+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\frac{(5x+2)^{4}}{4}+c\)
\(=\frac{1}{20}(5x+2)^{4}+c\)
\(Q.1.(xiii)\) \(\int{(2-7x)^4dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{35}(2-7x)^5+c\)
উত্তরঃ \(-\frac{1}{35}(2-7x)^5+c\)
সমাধানঃ
ধরি,
\(2-7x=t\)
\(\Rightarrow \frac{d}{dx}(2-7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-7.1=\frac{dt}{dx}\)
\(\Rightarrow -7=\frac{dt}{dx}\)
\(\Rightarrow -7dx=dt\)
\(\therefore dx=-\frac{1}{7}dt\)
\(\int{(2-7x)^4dx}\)\(2-7x=t\)
\(\Rightarrow \frac{d}{dx}(2-7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-7.1=\frac{dt}{dx}\)
\(\Rightarrow -7=\frac{dt}{dx}\)
\(\Rightarrow -7dx=dt\)
\(\therefore dx=-\frac{1}{7}dt\)
\(=\int{t^4.\left(-\frac{1}{7}dt\right)}\)
\(=-\frac{1}{7}\int{t^4dt}\)
\(=-\frac{1}{7}\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{7}\frac{t^{5}}{5}+c\)
\(=-\frac{1}{35}t^{5}+c\)
\(=-\frac{1}{35}(2-7x)^{5}+c\) ➜\(\because t=2-7x\)
বিকল্প পদ্ধতিঃ
\(\int{(2-7x)^4dx}\)
\(=\frac{1}{-7}\frac{(2-7x)^{4+1}}{4+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{7}\frac{(2-7x)^{5}}{5}+c\)
\(=-\frac{1}{35}(2-7x)^{5}+c\)
\(=\frac{1}{-7}\frac{(2-7x)^{4+1}}{4+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{7}\frac{(2-7x)^{5}}{5}+c\)
\(=-\frac{1}{35}(2-7x)^{5}+c\)
\(Q.1.(xiv)\) \(\int{\sqrt{1-x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
উত্তরঃ \(-\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
সমাধানঃ
ধরি,
\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(\int{\sqrt{1-x}dx}\)\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(=\int{(1-x)^{\frac{1}{2}}dx}\)
\(=\int{t^{\frac{1}{2}}(-dt)}\)
\(=-\int{t^{\frac{1}{2}}dt}\)
\(=-\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(1-x)^{\frac{3}{2}}+c\) ➜\(\because t=1-x\)
বিকল্প পদ্ধতিঃ
\(\int{\sqrt{1-x}dx}\)
\(=\int{(1-x)^{\frac{1}{2}}dx}\)
\(=\frac{1}{-1}\frac{(1-x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(1-x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\)
\(=-\frac{(1-x)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{(1-x)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(=\int{(1-x)^{\frac{1}{2}}dx}\)
\(=\frac{1}{-1}\frac{(1-x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(1-x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\)
\(=-\frac{(1-x)^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{(1-x)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(1-x)^{\frac{3}{2}}+c\)
\(Q.1.(xv)\) \(\int{\frac{dx}{\sqrt{2-3x}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{2}{3}\sqrt{2-3x}+c\)
উত্তরঃ \(-\frac{2}{3}\sqrt{2-3x}+c\)
সমাধানঃ
ধরি,
\(2-3x=t\)
\(\Rightarrow \frac{d}{dx}(2-3x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-3.1=\frac{dt}{dx}\)
\(\Rightarrow -3=\frac{dt}{dx}\)
\(\Rightarrow -3dx=dt\)
\(\therefore dx=-\frac{1}{3}dt\)
\(\int{\frac{dx}{\sqrt{2-3x}}}\)\(2-3x=t\)
\(\Rightarrow \frac{d}{dx}(2-3x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-3.1=\frac{dt}{dx}\)
\(\Rightarrow -3=\frac{dt}{dx}\)
\(\Rightarrow -3dx=dt\)
\(\therefore dx=-\frac{1}{3}dt\)
\(=\int{\frac{1}{\sqrt{2-3x}}dx}\)
\(=\int{\frac{1}{(2-3x)^{\frac{1}{2}}}dx}\)
\(=\int{(2-3x)^{-\frac{1}{2}}dx}\)
\(=\int{t^{-\frac{1}{2}}\left(-\frac{1}{3}dt\right)}\)
\(=-\frac{1}{3}\int{t^{-\frac{1}{2}}dt}\)
\(=-\frac{1}{3}\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{3}\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=-\frac{1}{3}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=-\frac{1}{3}.2t^{\frac{1}{2}}+c\)
\(=-\frac{2}{3}(2-3x)^{\frac{1}{2}}+c\) ➜\(\because t=2-3x\)
\(=-\frac{2}{3}\sqrt{2-3x}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{dx}{\sqrt{2-3x}}}\)
\(=\int{\frac{1}{\sqrt{2-3x}}dx}\)
\(=\int{\frac{1}{(2-3x)^{\frac{1}{2}}}dx}\)
\(=\int{(2-3x)^{-\frac{1}{2}}dx}\)
\(=\frac{1}{-3}\frac{(2-3x)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{3}\frac{(2-3x)^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=-\frac{1}{3}\frac{(2-3x)^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=-\frac{1}{3}.2(2-3x)^{\frac{1}{2}}+c\)
\(=-\frac{2}{3}(2-3x)^{\frac{1}{2}}+c\)
\(=-\frac{2}{3}\sqrt{2-3x}+c\)
\(=\int{\frac{1}{\sqrt{2-3x}}dx}\)
\(=\int{\frac{1}{(2-3x)^{\frac{1}{2}}}dx}\)
\(=\int{(2-3x)^{-\frac{1}{2}}dx}\)
\(=\frac{1}{-3}\frac{(2-3x)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{3}\frac{(2-3x)^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=-\frac{1}{3}\frac{(2-3x)^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=-\frac{1}{3}.2(2-3x)^{\frac{1}{2}}+c\)
\(=-\frac{2}{3}(2-3x)^{\frac{1}{2}}+c\)
\(=-\frac{2}{3}\sqrt{2-3x}+c\)
\(Q.1.(xvi)\) \(\int{\frac{dx}{(1-x)^2}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{1-x}+c\)
উত্তরঃ \(\frac{1}{1-x}+c\)
সমাধানঃ
ধরি,
\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(\int{\frac{dx}{(1-x)^2}}\)\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(=\int{\frac{1}{(1-x)^2}dx}\)
\(=\int{(1-x)^{-2}dx}\)
\(=\int{t^{-2}(-dt)}\)
\(=-\int{t^{-2}dt}\)
\(=-\frac{t^{-2+1}}{-2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{-1}}{-1}+c\) \(=t^{-1}+c\) \(=\frac{1}{t}+c\) \(=\frac{1}{1-x}+c\) ➜\(\because t=1-x\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{dx}{(1-x)^2}}\)
\(=\int{\frac{1}{(1-x)^2}dx}\)
\(=\int{(1-x)^{-2}dx}\)
\(=\frac{1}{-1}\frac{(1-x)^{-2+1}}{-2+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(1-x)^{-1}}{-1}+c\)
\(=(1-x)^{-1}+c\)
\(=\frac{1}{1-x}+c\)
\(=\int{\frac{1}{(1-x)^2}dx}\)
\(=\int{(1-x)^{-2}dx}\)
\(=\frac{1}{-1}\frac{(1-x)^{-2+1}}{-2+1}+c\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(1-x)^{-1}}{-1}+c\)
\(=(1-x)^{-1}+c\)
\(=\frac{1}{1-x}+c\)
\(Q.1.(xvii)\) \(\int{\frac{2x+1}{2x+3}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(x-\ln{|2x+3|}+c\)
উত্তরঃ \(x-\ln{|2x+3|}+c\)
সমাধানঃ
ধরি,
\(2x+3=t\)
\(\Rightarrow \frac{d}{dx}(2x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1+0=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(\int{\frac{2x+1}{2x+3}dx}\)\(2x+3=t\)
\(\Rightarrow \frac{d}{dx}(2x+3)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1+0=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(=\int{\frac{2x+3-2}{2x+3}dx}\)
\(=\int{\left(\frac{2x+3}{2x+3}-2\frac{1}{2x+3}\right)dx}\)
\(=\int{\left(1-2\frac{1}{2x+3}\right)dx}\)
\(=\int{dx}-2\int{\frac{1}{2x+3}dx}\)
\(=\int{dx}-2\int{\frac{1}{t}.\frac{1}{2}dt}\)
\(=\int{dx}-2.\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=\int{dx}-\int{\frac{1}{t}dt}\)
\(=x-\ln{|t|}+c\) ➜ \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-\ln{|2x+3|}+c\) ➜\(\because t=2x+3\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{2x+1}{2x+3}dx}\)
\(=\int{\frac{2x+3-2}{2x+3}dx}\)
\(=\int{\left(\frac{2x+3}{2x+3}-2\frac{1}{2x+3}\right)dx}\)
\(=\int{\left(1-2\frac{1}{2x+3}\right)dx}\)
\(=\int{dx}-2\int{\frac{1}{2x+3}dx}\)
\(=x-2\frac{1}{2}\ln{|2x+3|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-\ln{|2x+3|}+c\)
\(=\int{\frac{2x+3-2}{2x+3}dx}\)
\(=\int{\left(\frac{2x+3}{2x+3}-2\frac{1}{2x+3}\right)dx}\)
\(=\int{\left(1-2\frac{1}{2x+3}\right)dx}\)
\(=\int{dx}-2\int{\frac{1}{2x+3}dx}\)
\(=x-2\frac{1}{2}\ln{|2x+3|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x-\ln{|2x+3|}+c\)
\(Q.1.(xviii)\) \(\int{5(3-x)^{-1}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-5\ln{|3-x|}+c\)
উত্তরঃ \(-5\ln{|3-x|}+c\)
সমাধানঃ
ধরি,
\(3-x=t\)
\(\Rightarrow \frac{d}{dx}(3-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(\int{5(3-x)^{-1}dx}\)\(3-x=t\)
\(\Rightarrow \frac{d}{dx}(3-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(=\int{\frac{5}{3-x}dx}\)
\(=5\int{\frac{1}{3-x}dx}\)
\(=5\int{\frac{1}{t}(-dt)}\)
\(=-5\int{\frac{1}{t}dt}\)
\(=-5\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-5\ln{|t|}+c\)
\(=-5\ln{|3-x|}+c\) ➜\(\because t=3-x\)
বিকল্প পদ্ধতিঃ
\(\int{5(3-x)^{-1}dx}\)
\(=\int{\frac{5}{3-x}dx}\)
\(=5\int{\frac{1}{3-x}dx}\)
\(=5.\frac{1}{-1}\ln{|3-x|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-5\ln{|3-x|}+c\)
\(=\int{\frac{5}{3-x}dx}\)
\(=5\int{\frac{1}{3-x}dx}\)
\(=5.\frac{1}{-1}\ln{|3-x|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-5\ln{|3-x|}+c\)
\(Q.1.(xix)\) \(\int{\left(\frac{3}{x-1}-\frac{4}{x-2}\right)dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{\left\{\frac{(|x-1|)^3}{(x-2)^4}\right\}}+c\)
উত্তরঃ \(\ln{\left\{\frac{(|x-1|)^3}{(x-2)^4}\right\}}+c\)
সমাধানঃ
ধরি,
\(x-1=t_{1}\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 1-0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\therefore dx=dt_{1}\)
আবার,
\(x-2=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x-2)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\therefore dx=dt_{2}\)
\(\int{\left(\frac{3}{x-1}-\frac{4}{x-2}\right)dx}\)\(x-1=t_{1}\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 1-0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\therefore dx=dt_{1}\)
আবার,
\(x-2=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x-2)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\therefore dx=dt_{2}\)
\(=\int{\frac{3}{x-1}dx}-\int{\frac{4}{x-2}dx}\)
\(=3\int{\frac{1}{x-1}dx}-4\int{\frac{1}{x-2}dx}\)
\(=3\int{\frac{1}{t_{1}}dt_{1}}-4\int{\frac{1}{t_{2}}dt_{2}}\)
\(=3\ln{|t_{1}|}-4\ln{|t_{2}|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{(|t_{1}|)^3}-\ln{(|t_{2}|)^4}+c\)
\(=\ln{(|x-1|)^3}-\ln{(|x-2|)^4}+c\) ➜\(\because t_{1}=x-1, t_{2}=x-2\)
\(=\ln{\left\{\frac{(|x-1|)^3}{(|x-2|)^4}\right\}}+c\)
\(=\ln{\left\{\frac{(|x-1|)^3}{(x-2)^4}\right\}}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\left(\frac{3}{x-1}-\frac{4}{x-2}\right)dx}\)
\(=\int{\frac{3}{x-1}dx}-\int{\frac{4}{x-2}dx}\)
\(=3\int{\frac{1}{x-1}dx}-4\int{\frac{1}{x-2}dx}\)
\(=3.\frac{1}{1}\ln{|x-1|}-4.\frac{1}{1}\ln{|x-2|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=3\ln{|x-1|}-4\ln{|x-2|}+c\)
\(=\ln{(|x-1|)^3}-\ln{(|x-2|)^4}+c\)
\(=\ln{(|x-1|)^3}-\ln{(|x-2|)^4}+c\)
\(=\ln{\left\{\frac{(|x-1|)^3}{(|x-2|)^4}\right\}}+c\)
\(=\ln{\left\{\frac{(|x-1|)^3}{(x-2)^4}\right\}}+c\)
\(=\int{\frac{3}{x-1}dx}-\int{\frac{4}{x-2}dx}\)
\(=3\int{\frac{1}{x-1}dx}-4\int{\frac{1}{x-2}dx}\)
\(=3.\frac{1}{1}\ln{|x-1|}-4.\frac{1}{1}\ln{|x-2|}+c\) ➜ \(\because \int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=3\ln{|x-1|}-4\ln{|x-2|}+c\)
\(=\ln{(|x-1|)^3}-\ln{(|x-2|)^4}+c\)
\(=\ln{(|x-1|)^3}-\ln{(|x-2|)^4}+c\)
\(=\ln{\left\{\frac{(|x-1|)^3}{(|x-2|)^4}\right\}}+c\)
\(=\ln{\left\{\frac{(|x-1|)^3}{(x-2)^4}\right\}}+c\)
অনুশীলনী \(10.B / Q.2\)-এর সংক্ষিপ্ত প্রশ্নসমুহ
যোজিত ফল নির্ণয় করঃ
\(Q.2.(i)\) \(\int{e^{3x+2}dx}\)
উত্তরঃ \(\frac{1}{3}e^{3x+2}+c\)
\(Q.2.(ii)\) \(\int{\frac{dt}{2e^{2t}}}\)
উত্তরঃ \(-\frac{1}{4e^{2t}}+c\)
\(Q.2.(iii)\) \(\int{a^{4x}dx}\)
উত্তরঃ \(\frac{a^{4x}}{4\ln{a}}+c\)
\(Q.2.(iv)\) \(\int{\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
উত্তরঃ \(2\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)+c\)
\(Q.2.(v)\) \(\int{\frac{(e^x+1)^2}{\sqrt{e^x}}dx}\)
উত্তরঃ \(\frac{2}{3}e^{\frac{3x}{2}}+4e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
[ চঃ ২০১৪; মাঃ ২০১০ ]
\(Q.2.(vi)\) \(\int{\frac{e^{5x}+e^{3x}}{e^x+e^{-x}}dx}\)
উত্তরঃ \(\frac{1}{4}e^{4x}+c\)
[ চঃ ২০০০; দিঃ ২০১০ ]
\(Q.2.(vii)\) \(\int{\sin{7x}dx}\)
উত্তরঃ \(-\frac{1}{7}\cos{7x}+c\)
[ সিঃ ২০০৫ ]
\(Q.2.(viii)\) \(\int{\cos{7x}dx}\)
উত্তরঃ \(\frac{1}{7}\sin{7x}+c\)
\(Q.2.(ix)\) \(\int{\sin{5x}dx}\)
উত্তরঃ \(-\frac{1}{5}\cos{5x}+c\)
[ মাঃ ২০১৫ ]
\(Q.2.(x)\) \(\int{\cos{\left(5x+\frac{\pi}{3}\right)}dx}\)
উত্তরঃ \(\frac{1}{5}\sin{\left(5x+\frac{\pi}{3}\right)}+c\)
\(Q.2.(xi)\) \(\int{\sqrt{1-\cos{4x}}dx}\)
উত্তরঃ \(-\frac{1}{\sqrt{2}}\cos{2x}+c\)
[ চঃ২০০৭ ]
\(Q.2.(xii)\) \(\int{\sqrt{1+\cos{x}}dx}\)
উত্তরঃ \(2\sqrt{2}\sin{\frac{x}{2}}+c\)
[ বুয়েটঃ২০০৪-২০০৫ ]
\(Q.2.(xiii)\) \(\int{\sqrt{1-\cos{x}}dx}\)
উত্তরঃ \(-2\sqrt{2}\cos{\frac{x}{2}}+c\)
\(Q.2.(xiv)\) \(\int{\tan^2{\frac{x}{2}}dx}\)
উত্তরঃ \(2\tan{\frac{x}{2}}-x+c\)
\(Q.2.(xv)\) \(\int{\sec^2{(3x+2)}dx}\)
উত্তরঃ \(\frac{1}{3}\tan{(3x+2)}+c\)
\(Q.2.(xvi)\) \(\int{\frac{1-\cos{5x}}{1+\cos{5x}}dx}\)
উত্তরঃ \(\frac{2}{5}\tan{\frac{5x}{2}}-x+c\)
[ সিঃ২০০২ ]
\(Q.2.(xvii)\) \(\int{cosec^2{5x} \ dx}\)
উত্তরঃ \(-\frac{1}{5}\cot{5x}+c\)
\(Q.2.(xviii)\) \(\int{\frac{1}{1+\cos{x}}dx}\)
উত্তরঃ \(\tan{\frac{x}{2}}+c\)
[ কুঃ২০১৩; রাঃ ২০১১, ২০১০; যঃ ২০১১ ]
\(Q.2.(xix)\) \(\int{3\cos{3x}\cos{x}dx}\)
উত্তরঃ \(\frac{3}{8}(\sin{4x}+2\sin{2x})+c\)
\(Q.2.(xx)\) \(\int{\sin{5x}\sin{3x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; বঃ ২০০৮; চঃ২০১২ ]
\(Q.2.(xxi)\) \(\int{\sin{3x}\cos{5x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\cos{2x}-\cos{8x})+c\)
[ সিঃ২০১২; দিঃ২০১২ ]
\(Q.2.(xxii)\) \(\int{7\sin{4x}\sin{2x}dx}\)
উত্তরঃ \(\frac{7}{12}(3\sin{2x}-\sin{6x})+c\)
[ ঢাঃ২০১১; রাঃ২০০৫; যঃ২০০৪ ]
\(Q.2.(xxiii)\) \(\int{5\cos{4x}\sin{3x}dx}\)
উত্তরঃ \(\frac{5}{14}(7\cos{x}-\cos{7x})+c\)
[ সিঃ,দিঃ, রাঃ২০০৯; চঃ২০১১; কুঃ২০১৩; যঃ২০১২,২০১৪ ]
উত্তরঃ \(\frac{1}{3}e^{3x+2}+c\)
\(Q.2.(ii)\) \(\int{\frac{dt}{2e^{2t}}}\)
উত্তরঃ \(-\frac{1}{4e^{2t}}+c\)
\(Q.2.(iii)\) \(\int{a^{4x}dx}\)
উত্তরঃ \(\frac{a^{4x}}{4\ln{a}}+c\)
\(Q.2.(iv)\) \(\int{\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
উত্তরঃ \(2\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)+c\)
\(Q.2.(v)\) \(\int{\frac{(e^x+1)^2}{\sqrt{e^x}}dx}\)
উত্তরঃ \(\frac{2}{3}e^{\frac{3x}{2}}+4e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
[ চঃ ২০১৪; মাঃ ২০১০ ]
\(Q.2.(vi)\) \(\int{\frac{e^{5x}+e^{3x}}{e^x+e^{-x}}dx}\)
উত্তরঃ \(\frac{1}{4}e^{4x}+c\)
[ চঃ ২০০০; দিঃ ২০১০ ]
\(Q.2.(vii)\) \(\int{\sin{7x}dx}\)
উত্তরঃ \(-\frac{1}{7}\cos{7x}+c\)
[ সিঃ ২০০৫ ]
\(Q.2.(viii)\) \(\int{\cos{7x}dx}\)
উত্তরঃ \(\frac{1}{7}\sin{7x}+c\)
\(Q.2.(ix)\) \(\int{\sin{5x}dx}\)
উত্তরঃ \(-\frac{1}{5}\cos{5x}+c\)
[ মাঃ ২০১৫ ]
\(Q.2.(x)\) \(\int{\cos{\left(5x+\frac{\pi}{3}\right)}dx}\)
উত্তরঃ \(\frac{1}{5}\sin{\left(5x+\frac{\pi}{3}\right)}+c\)
\(Q.2.(xi)\) \(\int{\sqrt{1-\cos{4x}}dx}\)
উত্তরঃ \(-\frac{1}{\sqrt{2}}\cos{2x}+c\)
[ চঃ২০০৭ ]
\(Q.2.(xii)\) \(\int{\sqrt{1+\cos{x}}dx}\)
উত্তরঃ \(2\sqrt{2}\sin{\frac{x}{2}}+c\)
[ বুয়েটঃ২০০৪-২০০৫ ]
\(Q.2.(xiii)\) \(\int{\sqrt{1-\cos{x}}dx}\)
উত্তরঃ \(-2\sqrt{2}\cos{\frac{x}{2}}+c\)
\(Q.2.(xiv)\) \(\int{\tan^2{\frac{x}{2}}dx}\)
উত্তরঃ \(2\tan{\frac{x}{2}}-x+c\)
\(Q.2.(xv)\) \(\int{\sec^2{(3x+2)}dx}\)
উত্তরঃ \(\frac{1}{3}\tan{(3x+2)}+c\)
\(Q.2.(xvi)\) \(\int{\frac{1-\cos{5x}}{1+\cos{5x}}dx}\)
উত্তরঃ \(\frac{2}{5}\tan{\frac{5x}{2}}-x+c\)
[ সিঃ২০০২ ]
\(Q.2.(xvii)\) \(\int{cosec^2{5x} \ dx}\)
উত্তরঃ \(-\frac{1}{5}\cot{5x}+c\)
\(Q.2.(xviii)\) \(\int{\frac{1}{1+\cos{x}}dx}\)
উত্তরঃ \(\tan{\frac{x}{2}}+c\)
[ কুঃ২০১৩; রাঃ ২০১১, ২০১০; যঃ ২০১১ ]
\(Q.2.(xix)\) \(\int{3\cos{3x}\cos{x}dx}\)
উত্তরঃ \(\frac{3}{8}(\sin{4x}+2\sin{2x})+c\)
\(Q.2.(xx)\) \(\int{\sin{5x}\sin{3x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; বঃ ২০০৮; চঃ২০১২ ]
\(Q.2.(xxi)\) \(\int{\sin{3x}\cos{5x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\cos{2x}-\cos{8x})+c\)
[ সিঃ২০১২; দিঃ২০১২ ]
\(Q.2.(xxii)\) \(\int{7\sin{4x}\sin{2x}dx}\)
উত্তরঃ \(\frac{7}{12}(3\sin{2x}-\sin{6x})+c\)
[ ঢাঃ২০১১; রাঃ২০০৫; যঃ২০০৪ ]
\(Q.2.(xxiii)\) \(\int{5\cos{4x}\sin{3x}dx}\)
উত্তরঃ \(\frac{5}{14}(7\cos{x}-\cos{7x})+c\)
[ সিঃ,দিঃ, রাঃ২০০৯; চঃ২০১১; কুঃ২০১৩; যঃ২০১২,২০১৪ ]
\(Q.2.(xxiv)\) \(\int{4\cos{4x}\sin{5x}dx}\)
উত্তরঃ \(-\frac{2}{9}(\cos{9x}+9\cos{x})+c\)
[ রাঃ২০০৩ ]
\(Q.2.(xxv)\) \(\int{\sin{px}\cos{qx}dx}, (p>q)\)
উত্তরঃ \(-\frac{1}{2}\left\{\frac{\cos{(p+q)x}}{p+q}+\frac{\cos{(p-q)x}}{p-q}\right\}+c\)
[ সিঃ২০০৭; ঢাঃ২০০৩ ]
\(Q.2.(xxvi)\) \(\int{\sin{4x}\sin{2x}dx}\)
উত্তরঃ \(\frac{1}{12}(3\sin{2x}-\sin{6x})+c\)
\(Q.2.(xxvii)\) \(\int{3\sin{3x}\cos{4x}dx}\)
উত্তরঃ \(\frac{3}{14}(7\cos{x}-\cos{7x})+c\)
\(Q.2.(xxviii)\) \(\int{\cos^2{3\theta}d\theta}\)
উত্তরঃ \(\frac{1}{12}(6\theta-\sin{6\theta})+c\)
\(Q.2.(xxix)\) \(\int{\cos^2{\frac{\theta}{2}}d\theta}\)
উত্তরঃ \(\frac{1}{2}(\theta+\sin{\theta})+c\)
\(Q.2.(xxx)\) \(\int{\cos^2{2x}dx}\)
উত্তরঃ \(\frac{1}{8}(4x+\sin{4x})+c\)
[ ঢঃ২০০০ ]
\(Q.2.(xxxi)\) \(\int{\cos^2{\frac{x}{2}}dx}\)
উত্তরঃ \(\frac{1}{2}(x+\sin{x})+c\)
\(Q.2.(xxxii)\) \(\int{\sin^3{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{12}(\cos{3\theta}-9\cos{\theta})+c\)
\(Q.2.(xxxiii)\) \(\int{\cos^3{2x}dx}\)
উত্তরঃ \(\frac{1}{24}(\cos{6x}-9\cos{2x})+c\)
[ ঢাঃ২০০১]
\(Q.2.(xxxiv)\) \(\int{\sin^4{x}dx}\)
উত্তরঃ \(\frac{1}{32}(12x-8\sin{2x}+\sin{4x})+c\)
[ কুঃ২০০৯]
\(Q.2.(xxxv)\) \(\int{\left(1+\cos^2{\frac{x}{2}}\right)dx}\)
উত্তরঃ \(\frac{1}{2}(3x+\sin{x})+c\)
[ বুটেক্সঃ২০০৪-২০০৫ ]
\(Q.2.(xxxvi)\) \(\int{\sin^2{x}\cos{2x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{4x}-4x)+c\)
[ রাঃ২০১২,২০১৩; সিঃ২০১১; কুঃ২০০৭,২০১০,২০১৩,২০১৪; চঃ২০০২,২০০৯; যঃ২০০৫ ]
\(Q.2.(xxxvii)\) \(\int{(2\cos{x}+\sin{x})\cos{x}dx}\)
উত্তরঃ \(\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
[ ঢাঃ২০০৫ ]
\(Q.2.(xxxviii)\) \(\int{\sin^2{x}\cos^2{x}dx}\)
উত্তরঃ \(\frac{1}{32}(4x-\sin{4x})+c\)
[ ঢাঃ২০১৩; বঃ২০০৮; চঃ২০০৬; বুয়েটঃ২০০৩-২০০৪ ]
\(Q.2.(xxxix)\) \(\int{\cos^2{3\theta}d\theta}\)
উত্তরঃ \(\frac{1}{12}(6\theta+\sin{6\theta})+c\)
\(Q.2.(xL)\) \(\int{\cos^3{2x}dx}\)
উত্তরঃ \(\frac{1}{24}(\sin{6x}+9\sin{2x})+c\)
\(Q.2.(xLi)\) \(\int{\sqrt{1+\sin{x}}dx}\)
উত্তরঃ \(2(\sin{\frac{x}{2}}-\cos{\frac{x}{2}})+c\)
\(Q.2.(xLii)\) \(\int{\frac{1}{1+\sin{x}}dx}\)
উত্তরঃ \(\tan{x}-\sec{x}+c\)
[ চঃ২০১০; যঃ২০১৩]
\(Q.2.(xLiii)\) \(\int{\frac{1}{1-\cos{3x}}dx}\)
উত্তরঃ \(-\frac{1}{3}\cot{\frac{3x}{2}}+c\)
\(Q.2.(xLiv)\) \(\int{\frac{1+e^{5x}}{\sqrt{e^{3x}}}dx}\)
উত্তরঃ \(-\frac{2}{3}e^{-\frac{3x}{2}}+\frac{2}{7}e^{\frac{7x}{2}}+c\)
\(Q.2.(xLv)\) \(\int{\frac{1}{1-\sin{x}}dx}\)
উত্তরঃ \(\tan{x}+\sec{x}+c\)
উত্তরঃ \(-\frac{2}{9}(\cos{9x}+9\cos{x})+c\)
[ রাঃ২০০৩ ]
\(Q.2.(xxv)\) \(\int{\sin{px}\cos{qx}dx}, (p>q)\)
উত্তরঃ \(-\frac{1}{2}\left\{\frac{\cos{(p+q)x}}{p+q}+\frac{\cos{(p-q)x}}{p-q}\right\}+c\)
[ সিঃ২০০৭; ঢাঃ২০০৩ ]
\(Q.2.(xxvi)\) \(\int{\sin{4x}\sin{2x}dx}\)
উত্তরঃ \(\frac{1}{12}(3\sin{2x}-\sin{6x})+c\)
\(Q.2.(xxvii)\) \(\int{3\sin{3x}\cos{4x}dx}\)
উত্তরঃ \(\frac{3}{14}(7\cos{x}-\cos{7x})+c\)
\(Q.2.(xxviii)\) \(\int{\cos^2{3\theta}d\theta}\)
উত্তরঃ \(\frac{1}{12}(6\theta-\sin{6\theta})+c\)
\(Q.2.(xxix)\) \(\int{\cos^2{\frac{\theta}{2}}d\theta}\)
উত্তরঃ \(\frac{1}{2}(\theta+\sin{\theta})+c\)
\(Q.2.(xxx)\) \(\int{\cos^2{2x}dx}\)
উত্তরঃ \(\frac{1}{8}(4x+\sin{4x})+c\)
[ ঢঃ২০০০ ]
\(Q.2.(xxxi)\) \(\int{\cos^2{\frac{x}{2}}dx}\)
উত্তরঃ \(\frac{1}{2}(x+\sin{x})+c\)
\(Q.2.(xxxii)\) \(\int{\sin^3{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{12}(\cos{3\theta}-9\cos{\theta})+c\)
\(Q.2.(xxxiii)\) \(\int{\cos^3{2x}dx}\)
উত্তরঃ \(\frac{1}{24}(\cos{6x}-9\cos{2x})+c\)
[ ঢাঃ২০০১]
\(Q.2.(xxxiv)\) \(\int{\sin^4{x}dx}\)
উত্তরঃ \(\frac{1}{32}(12x-8\sin{2x}+\sin{4x})+c\)
[ কুঃ২০০৯]
\(Q.2.(xxxv)\) \(\int{\left(1+\cos^2{\frac{x}{2}}\right)dx}\)
উত্তরঃ \(\frac{1}{2}(3x+\sin{x})+c\)
[ বুটেক্সঃ২০০৪-২০০৫ ]
\(Q.2.(xxxvi)\) \(\int{\sin^2{x}\cos{2x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{4x}-4x)+c\)
[ রাঃ২০১২,২০১৩; সিঃ২০১১; কুঃ২০০৭,২০১০,২০১৩,২০১৪; চঃ২০০২,২০০৯; যঃ২০০৫ ]
\(Q.2.(xxxvii)\) \(\int{(2\cos{x}+\sin{x})\cos{x}dx}\)
উত্তরঃ \(\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
[ ঢাঃ২০০৫ ]
\(Q.2.(xxxviii)\) \(\int{\sin^2{x}\cos^2{x}dx}\)
উত্তরঃ \(\frac{1}{32}(4x-\sin{4x})+c\)
[ ঢাঃ২০১৩; বঃ২০০৮; চঃ২০০৬; বুয়েটঃ২০০৩-২০০৪ ]
\(Q.2.(xxxix)\) \(\int{\cos^2{3\theta}d\theta}\)
উত্তরঃ \(\frac{1}{12}(6\theta+\sin{6\theta})+c\)
\(Q.2.(xL)\) \(\int{\cos^3{2x}dx}\)
উত্তরঃ \(\frac{1}{24}(\sin{6x}+9\sin{2x})+c\)
\(Q.2.(xLi)\) \(\int{\sqrt{1+\sin{x}}dx}\)
উত্তরঃ \(2(\sin{\frac{x}{2}}-\cos{\frac{x}{2}})+c\)
\(Q.2.(xLii)\) \(\int{\frac{1}{1+\sin{x}}dx}\)
উত্তরঃ \(\tan{x}-\sec{x}+c\)
[ চঃ২০১০; যঃ২০১৩]
\(Q.2.(xLiii)\) \(\int{\frac{1}{1-\cos{3x}}dx}\)
উত্তরঃ \(-\frac{1}{3}\cot{\frac{3x}{2}}+c\)
\(Q.2.(xLiv)\) \(\int{\frac{1+e^{5x}}{\sqrt{e^{3x}}}dx}\)
উত্তরঃ \(-\frac{2}{3}e^{-\frac{3x}{2}}+\frac{2}{7}e^{\frac{7x}{2}}+c\)
\(Q.2.(xLv)\) \(\int{\frac{1}{1-\sin{x}}dx}\)
উত্তরঃ \(\tan{x}+\sec{x}+c\)
\(Q.2.(i)\) \(\int{e^{3x+2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}e^{3x+2}+c\)
উত্তরঃ \(\frac{1}{3}e^{3x+2}+c\)
সমাধানঃ
ধরি,
\(3x+2=t\)
\(\Rightarrow \frac{d}{dx}(3x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 3.1+0=\frac{dt}{dx}\)
\(\Rightarrow 3=\frac{dt}{dx}\)
\(\Rightarrow 3dx=dt\)
\(\therefore dx=\frac{1}{3}dt\)
\(\int{e^{3x+2}dx}\)\(3x+2=t\)
\(\Rightarrow \frac{d}{dx}(3x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 3.1+0=\frac{dt}{dx}\)
\(\Rightarrow 3=\frac{dt}{dx}\)
\(\Rightarrow 3dx=dt\)
\(\therefore dx=\frac{1}{3}dt\)
\(=\int{e^{t}.\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{e^{t}dt}\)
\(=\frac{1}{3}e^{t}+c\) ➜ \(\because \int{e^{x}dx}=e^{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}e^{3x+2}+c\) ➜\(\because t=3x+2\)
বিকল্প পদ্ধতিঃ
\(\int{e^{3x+2}dx}\)
\(=\frac{1}{3}e^{3x+2}+c\) ➜ \(\because \int{e^{ax+b}dx}=\frac{1}{a}e^{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}e^{3x+2}+c\) ➜ \(\because \int{e^{ax+b}dx}=\frac{1}{a}e^{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(ii)\) \(\int{\frac{dt}{2e^{2t}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{4e^{2t}}+c\)
উত্তরঃ \(-\frac{1}{4e^{2t}}+c\)
সমাধানঃ
ধরি,
\(-2t=x\)
\(\Rightarrow \frac{d}{dt}(-2x)=\frac{d}{dt}(x)\)
\(\Rightarrow -2.1=\frac{dx}{dt}\)
\(\Rightarrow -2=\frac{dx}{dt}\)
\(\Rightarrow -2dt=dx\)
\(\therefore dt=-\frac{1}{2}dx\)
\(\int{\frac{dt}{2e^{2t}}}\)\(-2t=x\)
\(\Rightarrow \frac{d}{dt}(-2x)=\frac{d}{dt}(x)\)
\(\Rightarrow -2.1=\frac{dx}{dt}\)
\(\Rightarrow -2=\frac{dx}{dt}\)
\(\Rightarrow -2dt=dx\)
\(\therefore dt=-\frac{1}{2}dx\)
\(=\frac{1}{2}\int{\frac{1}{e^{2t}}dt}\)
\(=\frac{1}{2}\int{e^{-2t}dt}\)
\(=\frac{1}{2}\int{e^{x}\times{-\frac{1}{2}dx}}\)
\(=-\frac{1}{4}\int{e^{x}dx}\)
\(=-\frac{1}{4}e^{x}+c\) ➜ \(\because \int{e^{x}dx}=e^{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}e^{-2t}+c\) ➜\(\because x=-2t\)
\(=-\frac{1}{4e^{2t}}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{dt}{2e^{2t}}}\)
\(=\frac{1}{2}\int{\frac{1}{e^{2t}}dt}\)
\(=\frac{1}{2}\int{e^{-2t}dt}\)
\(=\frac{1}{2}.\frac{1}{-2}e^{-2t}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}e^{-2t}+c\)
\(=-\frac{1}{4e^{2t}}+c\)
\(=\frac{1}{2}\int{\frac{1}{e^{2t}}dt}\)
\(=\frac{1}{2}\int{e^{-2t}dt}\)
\(=\frac{1}{2}.\frac{1}{-2}e^{-2t}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}e^{-2t}+c\)
\(=-\frac{1}{4e^{2t}}+c\)
\(Q.2.(iii)\) \(\int{a^{4x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{a^{4x}}{4\ln{a}}+c\)
উত্তরঃ \(\frac{a^{4x}}{4\ln{a}}+c\)
সমাধানঃ
ধরি,
\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(\int{a^{4x}dx}\)\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(=\int{a^{t}.\frac{1}{4}dt}\)
\(=\frac{1}{4}\int{a^{t}dt}\)
\(=\frac{1}{4}\frac{a^{t}}{\ln{a}}+c\) ➜ \(\because \int{a^{x}dx}=\frac{a^{x}}{\ln{a}}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\frac{a^{4x}}{\ln{a}}+c\) ➜\(\because t=4x\)
\(=\frac{a^{4x}}{4\ln{a}}+c\)
বিকল্প পদ্ধতিঃ
\(\int{a^{4x}dx}\)
\(=\frac{a^{4x}}{4\ln{a}}+c\) ➜ \(\because \int{a^{mx}dx}=\frac{a^{mx}}{m\ln{a}}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{a^{4x}}{4\ln{a}}+c\) ➜ \(\because \int{a^{mx}dx}=\frac{a^{mx}}{m\ln{a}}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(iv)\) \(\int{\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)+c\)
উত্তরঃ \(2\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)+c\)
সমাধানঃ
ধরি,
\(\frac{x}{2}=t_{1}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow \frac{1}{2}=\frac{dt_{1}}{dx}\)
\(\therefore dx=2dt_{1}\)
আবার,
\(-\frac{x}{2}=t_{2}\)
\(\Rightarrow \frac{d}{dx}\left(-\frac{x}{2}\right)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow -\frac{1}{2}=\frac{dt_{2}}{dx}\)
\(\Rightarrow -dx=2dt_{2}\)
\(\therefore dx=-2dt_{2}\)
\(\int{\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)\(\frac{x}{2}=t_{1}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow \frac{1}{2}=\frac{dt_{1}}{dx}\)
\(\therefore dx=2dt_{1}\)
আবার,
\(-\frac{x}{2}=t_{2}\)
\(\Rightarrow \frac{d}{dx}\left(-\frac{x}{2}\right)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow -\frac{1}{2}=\frac{dt_{2}}{dx}\)
\(\Rightarrow -dx=2dt_{2}\)
\(\therefore dx=-2dt_{2}\)
\(=\int{e^{\frac{x}{2}}dx}+\int{e^{-\frac{x}{2}}dx}\)
\(=\int{e^{t_{1}}.2dt_{1}}+\int{e^{t_{2}}(-2dt_{2})}\)
\(=2\int{e^{t_{1}}dt_{1}}-2\int{e^{t_{2}}dt_{2}}\)
\(=2e^{t_{1}}-2e^{t_{2}}+c\) ➜ \(\because \int{e^{x}dx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\) ➜\(\because t_{1}=\frac{x}{2}, t_{2}=-\frac{x}{2}\)
\(=2\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)+c\)
বিকল্প পদ্ধতিঃ
\(\int{\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{e^{\frac{x}{2}}dx}+\int{e^{-\frac{x}{2}}dx}\)
\(=\frac{1}{\frac{1}{2}}e^{\frac{x}{2}}+\frac{1}{-\frac{1}{2}}e^{-\frac{x}{2}}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
\(=2\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)+c\)
\(=\int{e^{\frac{x}{2}}dx}+\int{e^{-\frac{x}{2}}dx}\)
\(=\frac{1}{\frac{1}{2}}e^{\frac{x}{2}}+\frac{1}{-\frac{1}{2}}e^{-\frac{x}{2}}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
\(=2\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)+c\)
\(Q.2.(v)\) \(\int{\frac{(e^x+1)^2}{\sqrt{e^x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}e^{\frac{3x}{2}}+4e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
[ চঃ ২০১৪; মাঃ ২০১০ ]
উত্তরঃ \(\frac{2}{3}e^{\frac{3x}{2}}+4e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
[ চঃ ২০১৪; মাঃ ২০১০ ]
সমাধানঃ
ধরি,
\(\frac{3x}{2}=t_{1}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{3x}{2}\right)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow \frac{3}{2}=\frac{dt_{1}}{dx}\)
\(\Rightarrow \frac{3}{2}dx=dt_{1}\)
\(\therefore dx=\frac{2}{3}dt_{1}\)
আবার,
\(\frac{x}{2}=t_{2}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow \frac{1}{2}=\frac{dt_{2}}{dx}\)
\(\therefore dx=2dt_{2}\)
আবার,
\(-\frac{x}{2}=t_{3}\)
\(\Rightarrow \frac{d}{dx}\left(-\frac{x}{2}\right)=\frac{d}{dx}(t_{3})\)
\(\Rightarrow -\frac{1}{2}=\frac{dt_{3}}{dx}\)
\(\Rightarrow -dx=2dt_{3}\)
\(\therefore dx=-2dt_{3}\)
\(\int{\frac{(e^x+1)^2}{\sqrt{e^x}}dx}\)\(\frac{3x}{2}=t_{1}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{3x}{2}\right)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow \frac{3}{2}=\frac{dt_{1}}{dx}\)
\(\Rightarrow \frac{3}{2}dx=dt_{1}\)
\(\therefore dx=\frac{2}{3}dt_{1}\)
আবার,
\(\frac{x}{2}=t_{2}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow \frac{1}{2}=\frac{dt_{2}}{dx}\)
\(\therefore dx=2dt_{2}\)
আবার,
\(-\frac{x}{2}=t_{3}\)
\(\Rightarrow \frac{d}{dx}\left(-\frac{x}{2}\right)=\frac{d}{dx}(t_{3})\)
\(\Rightarrow -\frac{1}{2}=\frac{dt_{3}}{dx}\)
\(\Rightarrow -dx=2dt_{3}\)
\(\therefore dx=-2dt_{3}\)
\(=\int{\frac{(e^x)^2+2e^x+1}{\sqrt{e^x}}dx}\)
\(=\int{\frac{(e^x)^2+2e^x+1}{(e^x)^{\frac{1}{2}}}dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\int{\frac{e^{2x}+2e^x+1}{e^{\frac{x}{2}}}dx}\)
\(=\int{\left(\frac{e^{2x}}{e^{\frac{x}{2}}}+2\frac{e^x}{e^{\frac{x}{2}}}+\frac{1}{e^{\frac{x}{2}}}\right)dx}\)
\(=\int{\left(e^{2x-\frac{x}{2}}+2e^{x-\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{\left(e^{\frac{4x-x}{2}}+2e^{\frac{2x-x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{\left(e^{\frac{3x}{2}}+2e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{e^{\frac{3x}{2}}dx}+2\int{e^{\frac{x}{2}}dx}+\int{e^{-\frac{x}{2}}dx}\)
\(=\int{e^{t_{1}}.\frac{2}{3}dt_{1}}+2\int{e^{t_{2}}.2dt_{2}}+\int{e^{t_{3}}(-2dt_{3})}\)
\(=\frac{2}{3}\int{e^{t_{1}}dt_{1}}+4\int{e^{t_{2}}dt_{2}}-2\int{e^{t_{3}}dt_{3}}\)
\(=\frac{2}{3}e^{t_{1}}+4e^{t_{2}}-2e^{t_{3}}+c\) ➜ \(\because \int{e^{x}dx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{3}e^{\frac{3x}{2}}+4e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\) ➜\(\because t_{1}=\frac{3x}{2}, t_{2}=\frac{x}{2}, t_{3}=-\frac{x}{2}\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{(e^x+1)^2}{\sqrt{e^x}}dx}\)
\(=\int{\frac{(e^x)^2+2e^x+1}{\sqrt{e^x}}dx}\)
\(=\int{\frac{(e^x)^2+2e^x+1}{(e^x)^{\frac{1}{2}}}dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\int{\frac{e^{2x}+2e^x+1}{e^{\frac{x}{2}}}dx}\)
\(=\int{\left(\frac{e^{2x}}{e^{\frac{x}{2}}}+2\frac{e^x}{e^{\frac{x}{2}}}+\frac{1}{e^{\frac{x}{2}}}\right)dx}\)
\(=\int{\left(e^{2x-\frac{x}{2}}+2e^{x-\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{\left(e^{\frac{4x-x}{2}}+2e^{\frac{2x-x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{\left(e^{\frac{3x}{2}}+2e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{e^{\frac{3x}{2}}dx}+2\int{e^{\frac{x}{2}}dx}+\int{e^{-\frac{x}{2}}dx}\)
\(=\frac{1}{\frac{3}{2}}e^{\frac{3x}{2}}+2\frac{1}{\frac{1}{2}}e^{\frac{x}{2}}+\frac{1}{-\frac{1}{2}}e^{-\frac{x}{2}}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{3}e^{\frac{3x}{2}}+4e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
\(=\int{\frac{(e^x)^2+2e^x+1}{\sqrt{e^x}}dx}\)
\(=\int{\frac{(e^x)^2+2e^x+1}{(e^x)^{\frac{1}{2}}}dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\int{\frac{e^{2x}+2e^x+1}{e^{\frac{x}{2}}}dx}\)
\(=\int{\left(\frac{e^{2x}}{e^{\frac{x}{2}}}+2\frac{e^x}{e^{\frac{x}{2}}}+\frac{1}{e^{\frac{x}{2}}}\right)dx}\)
\(=\int{\left(e^{2x-\frac{x}{2}}+2e^{x-\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{\left(e^{\frac{4x-x}{2}}+2e^{\frac{2x-x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{\left(e^{\frac{3x}{2}}+2e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)dx}\)
\(=\int{e^{\frac{3x}{2}}dx}+2\int{e^{\frac{x}{2}}dx}+\int{e^{-\frac{x}{2}}dx}\)
\(=\frac{1}{\frac{3}{2}}e^{\frac{3x}{2}}+2\frac{1}{\frac{1}{2}}e^{\frac{x}{2}}+\frac{1}{-\frac{1}{2}}e^{-\frac{x}{2}}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{3}e^{\frac{3x}{2}}+4e^{\frac{x}{2}}-2e^{-\frac{x}{2}}+c\)
\(Q.2.(vi)\) \(\int{\frac{e^{5x}+e^{3x}}{e^x+e^{-x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}e^{4x}+c\)
[ চঃ ২০০০; দিঃ ২০১০ ]
উত্তরঃ \(\frac{1}{4}e^{4x}+c\)
[ চঃ ২০০০; দিঃ ২০১০ ]
সমাধানঃ
ধরি,
\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(\int{\frac{e^{5x}+e^{3x}}{e^x+e^{-x}}dx}\)\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(=\int{\frac{e^{4x}.e^{x}+e^{4x}.e^{-x}}{e^x+e^{-x}}dx}\)
\(=\int{\frac{e^{4x}(e^{x}+e^{-x})}{e^x+e^{-x}}dx}\)
\(=\int{e^{4x}dx}\)
\(=\int{e^{t}.\frac{1}{4}dt}\)
\(=\frac{1}{4}\int{e^{t}dt}\)
\(=\frac{1}{4}e^{t}+c\) ➜ \(\because \int{e^{x}dx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}e^{4x}+c\) ➜\(\because t=4x\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{e^{5x}+e^{3x}}{e^x+e^{-x}}dx}\)
\(=\int{\frac{e^{4x}.e^{x}+e^{4x}.e^{-x}}{e^x+e^{-x}}dx}\)
\(=\int{\frac{e^{4x}(e^{x}+e^{-x})}{e^x+e^{-x}}dx}\)
\(=\int{e^{4x}dx}\)
\(=\frac{1}{4}e^{4x}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\int{\frac{e^{4x}.e^{x}+e^{4x}.e^{-x}}{e^x+e^{-x}}dx}\)
\(=\int{\frac{e^{4x}(e^{x}+e^{-x})}{e^x+e^{-x}}dx}\)
\(=\int{e^{4x}dx}\)
\(=\frac{1}{4}e^{4x}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(vii)\) \(\int{\sin{7x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{7}\cos{7x}+c\)
[ সিঃ ২০০৫ ]
উত্তরঃ \(-\frac{1}{7}\cos{7x}+c\)
[ সিঃ ২০০৫ ]
সমাধানঃ
ধরি,
\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(\int{\sin{7x}dx}\)\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(=\int{\sin{t}.\frac{1}{7}dt}\)
\(=\frac{1}{7}\int{\sin{t}dt}\)
\(=\frac{1}{7}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{7}\cos{t}+c\)
\(=-\frac{1}{7}\cos{7x}+c\) ➜\(\because t=7x\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{7x}dx}\)
\(=-\frac{1}{7}\cos{7x}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{7}\cos{7x}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(viii)\) \(\int{\cos{7x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{7}\sin{7x}+c\)
উত্তরঃ \(\frac{1}{7}\sin{7x}+c\)
সমাধানঃ
ধরি,
\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(\int{\cos{7x}dx}\)\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(=\int{\sin{t}.\frac{1}{7}dt}\)
\(=\frac{1}{7}\int{\cos{t}dt}\)
\(=\frac{1}{7}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{7}\sin{t}+c\)
\(=\frac{1}{7}\sin{7x}+c\) ➜\(\because t=7x\)
বিকল্প পদ্ধতিঃ
\(\int{\cos{7x}dx}\)
\(=\frac{1}{7}\sin{7x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{7}\sin{7x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(ix)\) \(\int{\sin{5x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{5}\cos{5x}+c\)
[ মাঃ ২০১৫ ]
উত্তরঃ \(-\frac{1}{5}\cos{5x}+c\)
[ মাঃ ২০১৫ ]
সমাধানঃ
ধরি,
\(5x=t\)
\(\Rightarrow \frac{d}{dx}(5x)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(\int{\sin{5x}dx}\)\(5x=t\)
\(\Rightarrow \frac{d}{dx}(5x)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(=\int{\sin{t}.\frac{1}{5}dt}\)
\(=\frac{1}{5}\int{\sin{t}dt}\)
\(=\frac{1}{5}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{5}\cos{t}+c\)
\(=-\frac{1}{5}\cos{5x}+c\) ➜\(\because t=5x\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{5x}dx}\)
\(=-\frac{1}{5}\cos{5x}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{5}\cos{5x}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(x)\) \(\int{\cos{\left(5x+\frac{\pi}{3}\right)}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{5}\sin{\left(5x+\frac{\pi}{3}\right)}+c\)
উত্তরঃ \(\frac{1}{5}\sin{\left(5x+\frac{\pi}{3}\right)}+c\)
সমাধানঃ
ধরি,
\(5x+\frac{\pi}{3}=t\)
\(\Rightarrow \frac{d}{dx}\left(5x+\frac{\pi}{3}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1+0=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(\int{\cos{\left(5x+\frac{\pi}{3}\right)}dx}\)\(5x+\frac{\pi}{3}=t\)
\(\Rightarrow \frac{d}{dx}\left(5x+\frac{\pi}{3}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1+0=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(=\int{\cos{t}.\frac{1}{5}dt}\)
\(=\frac{1}{5}\int{\cos{t}dt}\)
\(=\frac{1}{5}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\sin{t}+c\)
\(=\frac{1}{5}\sin{\left(5x+\frac{\pi}{3}\right)}+c\) ➜\(\because t=5x+\frac{\pi}{3}\)
বিকল্প পদ্ধতিঃ
\(\int{\cos{\left(5x+\frac{\pi}{3}\right)}dx}\)
\(=\frac{1}{5}\sin{\left(5x+\frac{\pi}{3}\right)}+c\) ➜ \(\because \int{\cos{(ax+b)}dx}=\frac{1}{a}\sin{(ax+b)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{5}\sin{\left(5x+\frac{\pi}{3}\right)}+c\) ➜ \(\because \int{\cos{(ax+b)}dx}=\frac{1}{a}\sin{(ax+b)}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(xi)\) \(\int{\sqrt{1-\cos{4x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{\sqrt{2}}\cos{2x}+c\)
[ চঃ২০০৭ ]
উত্তরঃ \(-\frac{1}{\sqrt{2}}\cos{2x}+c\)
[ চঃ২০০৭ ]
সমাধানঃ
ধরি,
\(2x=t\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(\int{\sqrt{1-\cos{4x}}dx}\)\(2x=t\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(=\int{\sqrt{2\sin^2{2x}}dx}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
\(=\sqrt{2}\int{\sin{2x}dx}\)
\(=\sqrt{2}\int{\sin{t}.\frac{1}{2}dt}\)
\(=\sqrt{2}.\frac{1}{2}\int{\sin{t}dt}\)
\(=\frac{\sqrt{2}}{2}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\sqrt{2}}{\sqrt{2}.\sqrt{2}}\cos{t}+c\)
\(=-\frac{1}{\sqrt{2}}\cos{2x}+c\) ➜\(\because t=2x\)
বিকল্প পদ্ধতিঃ
\(\int{\sqrt{1-\cos{4x}}dx}\)
\(=\int{\sqrt{2\sin^2{2x}}dx}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
\(=\sqrt{2}\int{\sin{2x}dx}\)
\(=\sqrt{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\sqrt{2}.\frac{1}{2}\cos{2x}+c\)
\(=-\frac{\sqrt{2}}{\sqrt{2}.\sqrt{2}}\cos{2x}+c\)
\(=-\frac{1}{\sqrt{2}}\cos{2x}+c\)
\(=\int{\sqrt{2\sin^2{2x}}dx}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
\(=\sqrt{2}\int{\sin{2x}dx}\)
\(=\sqrt{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\sqrt{2}.\frac{1}{2}\cos{2x}+c\)
\(=-\frac{\sqrt{2}}{\sqrt{2}.\sqrt{2}}\cos{2x}+c\)
\(=-\frac{1}{\sqrt{2}}\cos{2x}+c\)
\(Q.2.(xii)\) \(\int{\sqrt{1+\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{2}\sin{\frac{x}{2}}+c\)
[ বুয়েটঃ২০০৪-২০০৫ ]
উত্তরঃ \(2\sqrt{2}\sin{\frac{x}{2}}+c\)
[ বুয়েটঃ২০০৪-২০০৫ ]
সমাধানঃ
ধরি,
\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(\int{\sqrt{1+\cos{x}}dx}\)\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(=\int{\sqrt{2\cos^2{\frac{x}{2}}}dx}\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\sqrt{2}\int{\sqrt{\cos^2{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\cos{\frac{x}{2}}dx}\)
\(=\sqrt{2}\int{\cos{t}.2dt}\)
\(=2\sqrt{2}\int{\cos{t}dt}\)
\(=2\sqrt{2}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sqrt{2}\sin{\frac{x}{2}}+c\) ➜\(\because t=\frac{x}{2}\)
বিকল্প পদ্ধতিঃ
\(\int{\sqrt{1+\cos{x}}dx}\)
\(=\int{\sqrt{2\cos^2{\frac{x}{2}}}dx}\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\sqrt{2}\int{\sqrt{\cos^2{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\cos{\frac{x}{2}}dx}\)
\(=\sqrt{2}\frac{1}{\frac{1}{2}}\sin{\frac{x}{2}}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sqrt{2}\sin{\frac{x}{2}}+c\)
\(=\int{\sqrt{2\cos^2{\frac{x}{2}}}dx}\) ➜ \(\because 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\sqrt{2}\int{\sqrt{\cos^2{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\cos{\frac{x}{2}}dx}\)
\(=\sqrt{2}\frac{1}{\frac{1}{2}}\sin{\frac{x}{2}}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sqrt{2}\sin{\frac{x}{2}}+c\)
\(Q.2.(xiii)\) \(\int{\sqrt{1-\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-2\sqrt{2}\cos{\frac{x}{2}}+c\)
উত্তরঃ \(-2\sqrt{2}\cos{\frac{x}{2}}+c\)
সমাধানঃ
ধরি,
\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(\int{\sqrt{1-\cos{x}}dx}\)\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(=\int{\sqrt{2\sin^2{\frac{x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(=\sqrt{2}\int{\sqrt{\sin^2{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\sin{\frac{x}{2}}dx}\)
\(=\sqrt{2}\int{\sin{t}.2dt}\)
\(=2\sqrt{2}\int{\sin{t}dt}\)
\(=2\sqrt{2}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-2\sqrt{2}\cos{\frac{x}{2}}+c\) ➜\(\because t=\frac{x}{2}\)
বিকল্প পদ্ধতিঃ
\(\int{\sqrt{1-\cos{x}}dx}\)
\(=\int{\sqrt{2\sin^2{\frac{x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(=\sqrt{2}\int{\sqrt{\sin^2{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\sin{\frac{x}{2}}dx}\)
\(=\sqrt{2}\left(-\frac{1}{\frac{1}{2}}\cos{\frac{x}{2}}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-2\sqrt{2}\cos{\frac{x}{2}}+c\)
\(=\int{\sqrt{2\sin^2{\frac{x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(=\sqrt{2}\int{\sqrt{\sin^2{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\sin{\frac{x}{2}}dx}\)
\(=\sqrt{2}\left(-\frac{1}{\frac{1}{2}}\cos{\frac{x}{2}}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-2\sqrt{2}\cos{\frac{x}{2}}+c\)
\(Q.2.(xiv)\) \(\int{\tan^2{\frac{x}{2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\tan{\frac{x}{2}}-x+c\)
উত্তরঃ \(2\tan{\frac{x}{2}}-x+c\)
সমাধানঃ
ধরি,
\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(\int{\tan^2{\frac{x}{2}}dx}\)\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(=\int{\left(\sec^2{\frac{x}{2}}-1\right)dx}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int{\sec^2{\frac{x}{2}}dx}-\int{dx}\)
\(=\int{\sec^2{t}.2dt}-\int{dx}\)
\(=2\int{\sec^2{t}dt}-\int{dx}\)
\(=2\tan{t}-x+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\tan{\frac{x}{2}}-x+c\) ➜\(\because t=\frac{x}{2}\)
বিকল্প পদ্ধতিঃ
\(\int{\tan^2{\frac{x}{2}}dx}\)
\(=\int{\left(\sec^2{\frac{x}{2}}-1\right)dx}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int{\sec^2{\frac{x}{2}}dx}-\int{dx}\)
\(=\frac{1}{\frac{1}{2}}\tan{\frac{x}{2}}-x+c\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\tan{\frac{x}{2}}-x+c\)
\(=\int{\left(\sec^2{\frac{x}{2}}-1\right)dx}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int{\sec^2{\frac{x}{2}}dx}-\int{dx}\)
\(=\frac{1}{\frac{1}{2}}\tan{\frac{x}{2}}-x+c\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\tan{\frac{x}{2}}-x+c\)
\(Q.2.(xv)\) \(\int{\sec^2{(3x+2)}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\tan{(3x+2)}+c\)
উত্তরঃ \(\frac{1}{3}\tan{(3x+2)}+c\)
সমাধানঃ
ধরি,
\(3x+2=t\)
\(\Rightarrow \frac{d}{dx}(3x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 3.1+0=\frac{dt}{dx}\)
\(\Rightarrow 3=\frac{dt}{dx}\)
\(\therefore 3dx=dt\)
\(\therefore dx=\frac{1}{3}dt\)
\(\int{\sec^2{(3x+2)}dx}\)\(3x+2=t\)
\(\Rightarrow \frac{d}{dx}(3x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 3.1+0=\frac{dt}{dx}\)
\(\Rightarrow 3=\frac{dt}{dx}\)
\(\therefore 3dx=dt\)
\(\therefore dx=\frac{1}{3}dt\)
\(=\int{\sec^2{t}.\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{\sec^2{t}dt}\)
\(=\frac{1}{3}\tan{t}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\tan{3x+2}+c\) ➜\(\because t=3x+2\)
বিকল্প পদ্ধতিঃ
\(\int{\sec^2{(3x+2)}dx}\)
\(=\frac{1}{3}\tan{(3x+2)}+c\) ➜ \(\because \int{\sec^2{ax+b}dx}=\frac{1}{a}\tan{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\tan{(3x+2)}+c\) ➜ \(\because \int{\sec^2{ax+b}dx}=\frac{1}{a}\tan{ax+b}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(xvi)\) \(\int{\frac{1-\cos{5x}}{1+\cos{5x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{5}\tan{\frac{5x}{2}}-x+c\)
[ সিঃ২০০২ ]
উত্তরঃ \(\frac{2}{5}\tan{\frac{5x}{2}}-x+c\)
[ সিঃ২০০২ ]
সমাধানঃ
ধরি,
\(\frac{5x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{5x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{5}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{5}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{5}{2}dx=dt\)
\(\therefore dx=\frac{2}{5}dt\)
\(\int{\frac{1-\cos{5x}}{1+\cos{5x}}dx}\)\(\frac{5x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{5x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{5}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{5}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{5}{2}dx=dt\)
\(\therefore dx=\frac{2}{5}dt\)
\(=\int{\frac{2\sin^2{\frac{5x}{2}}}{2\cos^2{\frac{5x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\int{\frac{\sin^2{\frac{5x}{2}}}{\cos^2{\frac{5x}{2}}}dx}\)
\(=\int{\tan^2{\frac{5x}{2}}dx}\)
\(=\int{\left(\sec^2{\frac{5x}{2}}-1\right)dx}\)
\(=\int{\sec^2{\frac{5x}{2}}dx}-\int{dx}\)
\(=\int{\sec^2{t}.\frac{2}{5}dt}-\int{dx}\)
\(=\frac{2}{5}\int{\sec^2{t}dt}-\int{dx}\)
\(=\frac{2}{5}\tan{t}-x+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{5}\tan{\frac{5x}{2}}-x+c\) ➜\(\because t=\frac{5x}{2}\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1-\cos{5x}}{1+\cos{5x}}dx}\)
\(=\int{\frac{2\sin^2{\frac{5x}{2}}}{2\cos^2{\frac{5x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\int{\frac{\sin^2{\frac{5x}{2}}}{\cos^2{\frac{5x}{2}}}dx}\)
\(=\int{\tan^2{\frac{5x}{2}}dx}\)
\(=\int{\left(\sec^2{\frac{5x}{2}}-1\right)dx}\)
\(=\int{\sec^2{\frac{5x}{2}}dx}-\int{dx}\)
\(=\frac{1}{\frac{5}{2}}\tan{\frac{5x}{2}}-x+c\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{5}\tan{\frac{5x}{2}}-x+c\)
\(=\int{\frac{2\sin^2{\frac{5x}{2}}}{2\cos^2{\frac{5x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(=\int{\frac{\sin^2{\frac{5x}{2}}}{\cos^2{\frac{5x}{2}}}dx}\)
\(=\int{\tan^2{\frac{5x}{2}}dx}\)
\(=\int{\left(\sec^2{\frac{5x}{2}}-1\right)dx}\)
\(=\int{\sec^2{\frac{5x}{2}}dx}-\int{dx}\)
\(=\frac{1}{\frac{5}{2}}\tan{\frac{5x}{2}}-x+c\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{5}\tan{\frac{5x}{2}}-x+c\)
\(Q.2.(xvii)\) \(\int{cosec^2{5x} \ dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{5}\cot{5x}+c\)
উত্তরঃ \(-\frac{1}{5}\cot{5x}+c\)
সমাধানঃ
ধরি,
\(5x=t\)
\(\Rightarrow \frac{d}{dx}(5x)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(\int{cosec^2{5x} \ dx}\)\(5x=t\)
\(\Rightarrow \frac{d}{dx}(5x)=\frac{d}{dx}(t)\)
\(\Rightarrow 5.1=\frac{dt}{dx}\)
\(\Rightarrow 5=\frac{dt}{dx}\)
\(\Rightarrow 5dx=dt\)
\(\therefore dx=\frac{1}{5}dt\)
\(=\int{cosec^2{t} \ .\frac{1}{5}dt}\)
\(=\frac{1}{5}\int{cosec^2{t} \ dt}\)
\(=\frac{1}{5}(-\cot{t})+c\) ➜ \(\because \int{cosec^2{x} \ dx}=-\cot{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{5}\cot{t}+c\)
\(=-\frac{1}{5}\cot{5x}+c\) ➜\(\because t=5x\)
বিকল্প পদ্ধতিঃ
\(\int{cosec^2{5x} \ dx}\)
\(=-\frac{1}{5}\cot{5x}+c\) ➜ \(\because \int{cosec^2{ax} \ dx}=-\frac{1}{a}\cot{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{5}\cot{5x}+c\) ➜ \(\because \int{cosec^2{ax} \ dx}=-\frac{1}{a}\cot{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(xviii)\) \(\int{\frac{1}{1+\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\tan{\frac{x}{2}}+c\)
[ কুঃ২০১৩; রাঃ ২০১১, ২০১০; যঃ ২০১১ ]
উত্তরঃ \(\tan{\frac{x}{2}}+c\)
[ কুঃ২০১৩; রাঃ ২০১১, ২০১০; যঃ ২০১১ ]
সমাধানঃ
\(\int{\frac{1}{1+\cos{x}}dx}\)
\(=\int{\frac{1-\cos{x}}{(1+\cos{x})(1-\cos{x})}dx}\) ➜ লব ও হরের সহিত \((1-\cos{x})\) গুণ করে।
\(=\int{\frac{1-\cos{x}}{1-\cos^2{x}}dx}\) ➜ \(\because a^2-b^2=(a-b)(a+b)\)
\(=\int{\frac{1-\cos{x}}{\sin^2{x}}dx}\) ➜ \(\because 1-\cos^2{A}=\sin^2{A}\)
\(=\int{\left(\frac{1}{\sin^2{x}}-\frac{\cos{x}}{\sin^2{x}}\right)dx}\)
\(=\int{\left(cosec^2{x}-\frac{1}{\sin{x}}.\frac{\cos{x}}{\sin{x}}\right)dx}\)
\(=\int{\left(cosec^2{x}-cosec \ {x}\cot{x}\right)dx}\)
\(=\int{cosec^2{x} \ dx}-\int{cosec \ {x}\cot{x}dx}\)
\(=-\cot{x}+cosec \ {x}+c\) ➜ \(\because \int{cosec^2{x} \ dx}=-\cot{x}, \int{cosec \ {x}\cot{x}dx}=-cosec \ {x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sin{x}}-\frac{\cos{x}}{\sin{x}}+c\) \(=\frac{1-\cos{x}}{\sin{x}}+c\) \(=\frac{2\sin^2{\frac{x}{2}}}{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}+c\) ➜\(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(=\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}+c\)
\(=\tan{\frac{x}{2}}+c\)
\(=\int{\frac{1-\cos{x}}{(1+\cos{x})(1-\cos{x})}dx}\) ➜ লব ও হরের সহিত \((1-\cos{x})\) গুণ করে।
\(=\int{\frac{1-\cos{x}}{1-\cos^2{x}}dx}\) ➜ \(\because a^2-b^2=(a-b)(a+b)\)
\(=\int{\frac{1-\cos{x}}{\sin^2{x}}dx}\) ➜ \(\because 1-\cos^2{A}=\sin^2{A}\)
\(=\int{\left(\frac{1}{\sin^2{x}}-\frac{\cos{x}}{\sin^2{x}}\right)dx}\)
\(=\int{\left(cosec^2{x}-\frac{1}{\sin{x}}.\frac{\cos{x}}{\sin{x}}\right)dx}\)
\(=\int{\left(cosec^2{x}-cosec \ {x}\cot{x}\right)dx}\)
\(=\int{cosec^2{x} \ dx}-\int{cosec \ {x}\cot{x}dx}\)
\(=-\cot{x}+cosec \ {x}+c\) ➜ \(\because \int{cosec^2{x} \ dx}=-\cot{x}, \int{cosec \ {x}\cot{x}dx}=-cosec \ {x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{\sin{x}}-\frac{\cos{x}}{\sin{x}}+c\) \(=\frac{1-\cos{x}}{\sin{x}}+c\) \(=\frac{2\sin^2{\frac{x}{2}}}{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}+c\) ➜\(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(=\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}+c\)
\(=\tan{\frac{x}{2}}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1}{1+\cos{x}}dx}\)
\(=\int{\frac{1}{2\cos^2{\frac{x}{2}}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\cos^2{\frac{x}{2}}}dx}\)
\(=\frac{1}{2}\int{\sec^2{\frac{x}{2}}dx}\)
\(=\frac{1}{2}.\frac{1}{\frac{1}{2}}\tan{\frac{x}{2}}+c\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan{\frac{x}{2}}+c\)
\(=\int{\frac{1}{2\cos^2{\frac{x}{2}}}dx}\)
\(=\frac{1}{2}\int{\frac{1}{\cos^2{\frac{x}{2}}}dx}\)
\(=\frac{1}{2}\int{\sec^2{\frac{x}{2}}dx}\)
\(=\frac{1}{2}.\frac{1}{\frac{1}{2}}\tan{\frac{x}{2}}+c\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan{\frac{x}{2}}+c\)
\(Q.2.(xix)\) \(\int{3\cos{3x}\cos{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{3}{8}(\sin{4x}+2\sin{2x})+c\)
উত্তরঃ \(\frac{3}{8}(\sin{4x}+2\sin{2x})+c\)
সমাধানঃ
ধরি,
\(4x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 4.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4dx=dt_{1}\)
\(\therefore dx=\frac{1}{4}dt_{1}\)
আবার,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(\int{3\cos{3x}\cos{x}dx}\)\(4x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 4.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4dx=dt_{1}\)
\(\therefore dx=\frac{1}{4}dt_{1}\)
আবার,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(=\frac{3}{2}\int{2\cos{3x}\cos{x}dx}\)
\(=\frac{3}{2}\int{\{\cos{3x+x}+\cos{3x-x}\}dx}\) ➜\(2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}\)
\(=\frac{3}{2}\int{\{\cos{4x}+\cos{2x}\}dx}\)
\(=\frac{3}{2}\int{\cos{4x}dx}+\frac{3}{2}\int{\cos{2x}dx}\)
\(=\frac{3}{2}\int{\cos{t_{1}}.\frac{1}{4}dt_{1}}+\frac{3}{2}\int{\cos{t_{2}}.\frac{1}{2}dt_{2}}\)
\(=\frac{3}{8}\int{\cos{t_{1}}dt_{1}}+\frac{3}{4}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{3}{8}\sin{t_{1}}+\frac{3}{4}\sin{t_{2}}+c\)| \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{3}{8}\sin{4x}+\frac{3}{4}\sin{2x}+c\) ➜ \(\because t_{1}=4x, t_{2}=2x\)
\(=\frac{3}{8}(\sin{4x}+2\sin{2x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{3\cos{3x}\cos{x}dx}\)
\(=\frac{3}{2}\int{2\cos{3x}\cos{x}dx}\)
\(=\frac{3}{2}\int{\{\cos{3x+x}+\cos{3x-x}\}dx}\) ➜\(2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}\)
\(=\frac{3}{2}\int{\{\cos{4x}+\cos{2x}\}dx}\)
\(=\frac{3}{2}\int{\cos{4x}dx}+\frac{3}{2}\int{\cos{2x}dx}\)
\(=\frac{3}{2}.\frac{1}{4}\sin{4x}+\frac{3}{2}.\frac{1}{2}\sin{2x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{3}{8}\sin{4x}+\frac{3}{4}\sin{2x}+c\)
\(=\frac{3}{8}(\sin{4x}+2\sin{2x})+c\)
\(=\frac{3}{2}\int{2\cos{3x}\cos{x}dx}\)
\(=\frac{3}{2}\int{\{\cos{3x+x}+\cos{3x-x}\}dx}\) ➜\(2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}\)
\(=\frac{3}{2}\int{\{\cos{4x}+\cos{2x}\}dx}\)
\(=\frac{3}{2}\int{\cos{4x}dx}+\frac{3}{2}\int{\cos{2x}dx}\)
\(=\frac{3}{2}.\frac{1}{4}\sin{4x}+\frac{3}{2}.\frac{1}{2}\sin{2x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{3}{8}\sin{4x}+\frac{3}{4}\sin{2x}+c\)
\(=\frac{3}{8}(\sin{4x}+2\sin{2x})+c\)
\(Q.2.(xx)\) \(\int{\sin{5x}\sin{3x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; বঃ ২০০৮; চঃ২০১২ ]
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
[ যঃ২০১০; বঃ ২০০৮; চঃ২০১২ ]
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(8x=t_{2}\) \(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 8.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8dx=dt_{2}\)
\(\therefore dx=\frac{1}{8}dt_{2}\)
\(\int{\sin{5x}\sin{3x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(8x=t_{2}\) \(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 8.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8dx=dt_{2}\)
\(\therefore dx=\frac{1}{8}dt_{2}\)
\(=\frac{1}{2}\int{2\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(5x-3x)}-\cos{(5x+3x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos{8x}dx}\)
\(=\frac{1}{2}\int{\cos{t_{1}}.\frac{1}{2}dt_{1}}-\frac{1}{2}\int{\cos{t_{2}}.\frac{1}{8}dt_{2}}\)
\(=\frac{1}{4}\int{\cos{t_{1}}dt_{1}}-\frac{1}{16}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{4}\sin{t_{1}}+\frac{1}{16}\sin{t_{2}}+c\)| \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{16}\sin{8x}+c\) ➜ \(\because t_{1}=2x, t_{2}=8x\)
\(=\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{2\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(5x-3x)}-\cos{(5x+3x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos{8x}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{2}.\frac{1}{8}\sin{8x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{16}\sin{8x}+c\)
\(=\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
\(=\frac{1}{2}\int{2\sin{5x}\sin{3x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(5x-3x)}-\cos{(5x+3x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos{8x}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{2}.\frac{1}{8}\sin{8x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{16}\sin{8x}+c\)
\(=\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
\(Q.2.(xxi)\) \(\int{\sin{3x}\cos{5x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{16}(4\cos{2x}-\cos{8x})+c\)
[ সিঃ২০১২; দিঃ২০১২ ]
উত্তরঃ \(\frac{1}{16}(4\cos{2x}-\cos{8x})+c\)
[ সিঃ২০১২; দিঃ২০১২ ]
সমাধানঃ
ধরি,
\(8x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 8.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 8=\frac{dt_{1}}{dx}\)
\(\Rightarrow 8dx=dt_{1}\)
\(\therefore dx=\frac{1}{8}dt_{1}\)
আবার,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(\int{\sin{3x}\cos{5x}dx}\)\(8x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 8.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 8=\frac{dt_{1}}{dx}\)
\(\Rightarrow 8dx=dt_{1}\)
\(\therefore dx=\frac{1}{8}dt_{1}\)
আবার,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(=\frac{1}{2}\int{2\sin{3x}\cos{5x}dx}\)
\(=\frac{1}{2}\int{\{\sin{(3x+5x)}+\sin{(3x-5x)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int{\sin{8x}dx}+\frac{1}{2}\int{\sin{(-2x)}dx}\)
\(=\frac{1}{2}\int{\sin{8x}dx}-\frac{1}{2}\int{\sin{2x}dx}\)
\(=\frac{1}{2}\int{\sin{t_{1}}.\frac{1}{8}dt_{1}}-\frac{1}{2}\int{\sin{t_{2}}.\frac{1}{2}dt_{2}}\)
\(=\frac{1}{16}\int{\sin{t_{1}}dt_{1}}-\frac{1}{4}\int{\sin{t_{2}}dt_{2}}\)
\(=\frac{1}{16}(-\cos{t_{1}})-\frac{1}{4}(-\cos{t_{2}})+c\)| \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{16}\sin{8x}+\frac{1}{4}\sin{2x}+c\) ➜ \(\because t_{1}=8x, t_{2}=2x\)
\(=\frac{1}{16}(4\sin{2x}-\sin{8x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{3x}\cos{5x}dx}\)
\(=\frac{1}{2}\int{2\sin{3x}\cos{5x}dx}\)
\(=\frac{1}{2}\int{\{\sin{(3x+5x)}+\sin{(3x-5x)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int{\sin{8x}dx}+\frac{1}{2}\int{\sin{(-2x)}dx}\)
\(=\frac{1}{2}\int{\sin{8x}dx}-\frac{1}{2}\int{\sin{2x}dx}\)
\(=\frac{1}{2}\left\{-\frac{1}{8}\cos{8x}\right\}-\frac{1}{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{16}\cos{8x}+\frac{1}{4}\cos{2x}+c\)
\(=\frac{1}{16}(4\cos{2x}-\cos{8x})+c\)
\(=\frac{1}{2}\int{2\sin{3x}\cos{5x}dx}\)
\(=\frac{1}{2}\int{\{\sin{(3x+5x)}+\sin{(3x-5x)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int{\sin{8x}dx}+\frac{1}{2}\int{\sin{(-2x)}dx}\)
\(=\frac{1}{2}\int{\sin{8x}dx}-\frac{1}{2}\int{\sin{2x}dx}\)
\(=\frac{1}{2}\left\{-\frac{1}{8}\cos{8x}\right\}-\frac{1}{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{16}\cos{8x}+\frac{1}{4}\cos{2x}+c\)
\(=\frac{1}{16}(4\cos{2x}-\cos{8x})+c\)
\(Q.2.(xxii)\) \(\int{7\sin{4x}\sin{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{7}{12}(3\sin{2x}-\sin{6x})+c\)
[ ঢাঃ২০১১; রাঃ২০০৫; যঃ২০০৪ ]
উত্তরঃ \(\frac{7}{12}(3\sin{2x}-\sin{6x})+c\)
[ ঢাঃ২০১১; রাঃ২০০৫; যঃ২০০৪ ]
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(6x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 6.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6dx=dt_{2}\)
\(\therefore dx=\frac{1}{6}dt_{2}\)
\(\int{7\sin{4x}\sin{2x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(6x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 6.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6dx=dt_{2}\)
\(\therefore dx=\frac{1}{6}dt_{2}\)
\(=\frac{7}{2}\int{2\sin{4x}\sin{2x}dx}\)
\(=\frac{7}{2}\int{\{\cos{(4x-2x)}-\cos{(4x+2x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{7}{2}\int{\cos{2x}dx}-\frac{7}{2}\int{\cos{6x}dx}\)
\(=\frac{7}{2}\int{\cos{t_{1}}.\frac{1}{2}dt_{1}}-\frac{7}{2}\int{\cos{t_{2}}.\frac{1}{6}dt_{2}}\)
\(=\frac{7}{4}\int{\cos{t_{1}}dt_{1}}-\frac{7}{12}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{7}{4}\sin{t_{1}}-\frac{7}{12}\sin{t_{2}}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{7}{4}\sin{2x}-\frac{7}{12}\sin{6x}+c\) ➜ \(\because t_{1}=2x, t_{2}=6x\)
\(=\frac{7}{12}(3\sin{2x}-\sin{6x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{7\sin{4x}\sin{2x}dx}\)
\(=\frac{7}{2}\int{2\sin{4x}\sin{2x}dx}\)
\(=\frac{7}{2}\int{\{\cos{(4x-2x)}-\cos{(4x+2x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{7}{2}\int{\cos{2x}dx}-\frac{7}{2}\int{\cos{6x}dx}\)
\(=\frac{7}{2}.\frac{1}{2}\sin{2x}-\frac{7}{2}.\frac{1}{6}\sin{6x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{7}{4}\sin{2x}-\frac{7}{12}\sin{6x}+c\)
\(=\frac{7}{12}(3\sin{2x}-\sin{6x})+c\)
\(=\frac{7}{2}\int{2\sin{4x}\sin{2x}dx}\)
\(=\frac{7}{2}\int{\{\cos{(4x-2x)}-\cos{(4x+2x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{7}{2}\int{\cos{2x}dx}-\frac{7}{2}\int{\cos{6x}dx}\)
\(=\frac{7}{2}.\frac{1}{2}\sin{2x}-\frac{7}{2}.\frac{1}{6}\sin{6x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{7}{4}\sin{2x}-\frac{7}{12}\sin{6x}+c\)
\(=\frac{7}{12}(3\sin{2x}-\sin{6x})+c\)
\(Q.2.(xxiii)\) \(\int{5\cos{4x}\sin{3x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{5}{14}(7\cos{x}-\cos{7x})+c\)
[ সিঃ,দিঃ, রাঃ২০০৯; চঃ২০১১; কুঃ২০১৩; যঃ২০১২,২০১৪ ]
উত্তরঃ \(\frac{5}{14}(7\cos{x}-\cos{7x})+c\)
[ সিঃ,দিঃ, রাঃ২০০৯; চঃ২০১১; কুঃ২০১৩; যঃ২০১২,২০১৪ ]
সমাধানঃ
ধরি,
\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(\int{5\cos{4x}\sin{3x}dx}\)\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(=\frac{5}{2}\int{2\cos{4x}\sin{3x}dx}\)
\(=\frac{5}{2}\int{\{\sin{(4x+3x)}-\sin{(4x-3x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=\frac{5}{2}\int{\sin{7x}dx}-\frac{5}{2}\int{\sin{x}dx}\)
\(=\frac{5}{2}\int{\sin{t}.\frac{1}{7}dt}-\frac{5}{2}\int{\sin{x}dx}\)
\(=\frac{5}{14}\int{\sin{t}dt}-\frac{5}{2}\int{\sin{x}dx}\)
\(=\frac{5}{14}(-\cos{t})-\frac{5}{2}(-\cos{x})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{5}{14}\cos{7x}+\frac{5}{2}\cos{x}+c\) ➜ \(\because t=7x\)
\(=\frac{5}{14}(7\cos{x}-\cos{7x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{5\cos{4x}\sin{3x}dx}\)
\(=\frac{5}{2}\int{2\cos{4x}\sin{3x}dx}\)
\(=\frac{5}{2}\int{\{\sin{(4x+3x)}-\sin{(4x-3x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=\frac{5}{2}\int{\sin{7x}dx}-\frac{5}{2}\int{\sin{x}dx}\)
\(=\frac{5}{2}\left\{-\frac{1}{7}\cos{7x}\right\}-\frac{5}{2}(-\cos{x})+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{5}{14}\cos{7x}+\frac{5}{2}\cos{x}+c\)
\(=\frac{5}{14}(7\cos{x}-\cos{7x})+c\)
\(=\frac{5}{2}\int{2\cos{4x}\sin{3x}dx}\)
\(=\frac{5}{2}\int{\{\sin{(4x+3x)}-\sin{(4x-3x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=\frac{5}{2}\int{\sin{7x}dx}-\frac{5}{2}\int{\sin{x}dx}\)
\(=\frac{5}{2}\left\{-\frac{1}{7}\cos{7x}\right\}-\frac{5}{2}(-\cos{x})+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{5}{14}\cos{7x}+\frac{5}{2}\cos{x}+c\)
\(=\frac{5}{14}(7\cos{x}-\cos{7x})+c\)
\(Q.2.(xxiv)\) \(\int{4\cos{4x}\sin{5x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{2}{9}(\cos{9x}+9\cos{x})+c\)
[ রাঃ২০০৩ ]
উত্তরঃ \(-\frac{2}{9}(\cos{9x}+9\cos{x})+c\)
[ রাঃ২০০৩ ]
সমাধানঃ
ধরি,
\(9x=t\)
\(\Rightarrow \frac{d}{dx}(9x)=\frac{d}{dx}(t)\)
\(\Rightarrow 9.1=\frac{dt}{dx}\)
\(\Rightarrow 9=\frac{dt}{dx}\)
\(\Rightarrow 9dx=dt\)
\(\therefore dx=\frac{1}{9}dt\)
\(\int{4\cos{4x}\sin{5x}dx}\)\(9x=t\)
\(\Rightarrow \frac{d}{dx}(9x)=\frac{d}{dx}(t)\)
\(\Rightarrow 9.1=\frac{dt}{dx}\)
\(\Rightarrow 9=\frac{dt}{dx}\)
\(\Rightarrow 9dx=dt\)
\(\therefore dx=\frac{1}{9}dt\)
\(=2\int{2\cos{4x}\sin{5x}dx}\)
\(=2\int{\{\sin{(4x+5x)}-\sin{(4x-5x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=2\int{\{\sin{(9x)}-\sin{(-x)}\}dx}\)
\(=2\int{\sin{9x}dx}+2\int{\sin{x}dx}\)
\(=2\int{\sin{t}.\frac{1}{9}dt}+2\int{\sin{x}dx}\)
\(=\frac{2}{9}\int{\sin{t}dt}+2\int{\sin{x}dx}\)
\(=\frac{2}{9}(-\cos{t})+2(-\cos{x})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{9}\cos{9x}-2\cos{x}+c\) ➜ \(\because t=9x\)
\(=-\frac{2}{9}(\cos{9x}+9\cos{x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{4\cos{4x}\sin{5x}dx}\)
\(=2\int{2\cos{4x}\sin{5x}dx}\)
\(=2\int{\{\sin{(4x+5x)}-\sin{(4x-5x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=2\int{\{\sin{(9x)}-\sin{(-x)}\}dx}\)
\(=2\int{\sin{9x}dx}+2\int{\sin{x}dx}\)
\(=2\left\{-\frac{1}{9}\cos{9x}\right\}+2(-\cos{x})+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{9}\cos{9x}-2\cos{x}+c\)
\(=-\frac{2}{9}(\cos{9x}+9\cos{x})+c\)
\(=2\int{2\cos{4x}\sin{5x}dx}\)
\(=2\int{\{\sin{(4x+5x)}-\sin{(4x-5x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=2\int{\{\sin{(9x)}-\sin{(-x)}\}dx}\)
\(=2\int{\sin{9x}dx}+2\int{\sin{x}dx}\)
\(=2\left\{-\frac{1}{9}\cos{9x}\right\}+2(-\cos{x})+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{9}\cos{9x}-2\cos{x}+c\)
\(=-\frac{2}{9}(\cos{9x}+9\cos{x})+c\)
\(Q.2.(xxv)\) \(\int{\sin{px}\cos{qx}dx}, (p>q)\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{2}\left\{\frac{\cos{(p+q)x}}{p+q}+\frac{\cos{(p-q)x}}{p-q}\right\}+c\)
[ সিঃ২০০৭; ঢাঃ২০০৩ ]
উত্তরঃ \(-\frac{1}{2}\left\{\frac{\cos{(p+q)x}}{p+q}+\frac{\cos{(p-q)x}}{p-q}\right\}+c\)
[ সিঃ২০০৭; ঢাঃ২০০৩ ]
সমাধানঃ
ধরি,
\((p+q)x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(p+q)x=\frac{d}{dx}(t_{1})\)
\(\Rightarrow (p+q).1=\frac{dt_{1}}{dx}\)
\(\Rightarrow (p+q)=\frac{dt_{1}}{dx}\)
\(\Rightarrow (p+q)dx=dt_{1}\)
\(\therefore dx=\frac{1}{p+q}dt_{1}\)
আবার,
\((p-q)x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(p-q)x=\frac{d}{dx}(t_{2})\)
\(\Rightarrow (p-q).1=\frac{dt_{2}}{dx}\)
\(\Rightarrow (p-q)=\frac{dt_{2}}{dx}\)
\(\Rightarrow (p-q)dx=dt_{2}\)
\(\therefore dx=\frac{1}{p-q}dt_{2}\)
\(\int{\sin{px}\cos{qx}dx}\)\((p+q)x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(p+q)x=\frac{d}{dx}(t_{1})\)
\(\Rightarrow (p+q).1=\frac{dt_{1}}{dx}\)
\(\Rightarrow (p+q)=\frac{dt_{1}}{dx}\)
\(\Rightarrow (p+q)dx=dt_{1}\)
\(\therefore dx=\frac{1}{p+q}dt_{1}\)
আবার,
\((p-q)x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(p-q)x=\frac{d}{dx}(t_{2})\)
\(\Rightarrow (p-q).1=\frac{dt_{2}}{dx}\)
\(\Rightarrow (p-q)=\frac{dt_{2}}{dx}\)
\(\Rightarrow (p-q)dx=dt_{2}\)
\(\therefore dx=\frac{1}{p-q}dt_{2}\)
\(=\frac{1}{2}\int{2\sin{px}\cos{qx}dx}\)
\(=\frac{1}{2}\int{\{\sin{(px+qx)}+\sin{(px-qx)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int{\sin{(p+q)x}dx}+\frac{1}{2}\int{\sin{(p-q)x}dx}\)
\(=\frac{1}{2}\int{\sin{t_{1}}.\frac{1}{p+q}dt_{1}}+\frac{1}{2}\int{\sin{t_{2}}.\frac{1}{p-q}dt_{2}}\)
\(=\frac{1}{2(p+q)}\int{\sin{t_{1}}dt_{1}}+\frac{1}{2(p-q)}\int{\sin{t_{2}}dt_{2}}\)
\(=\frac{1}{2(p+q)}(-\cos{t_{1}})+\frac{1}{2(p-q)}(-\sin{t_{2}})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2(p+q)}\cos{t_{1}}-\frac{1}{2(p-q)}\sin{t_{2}}+c\)
\(=-\frac{1}{2(p+q)}\cos{(p+q)x}-\frac{1}{2(p-q)}\sin{(p-q)x}+c\) ➜ \(\because t_{1}=(p+q)x, t_{2}=(p-q)x\)
\(=-\frac{1}{2}\left\{\frac{\cos{(p+q)x}}{p+q}+\frac{\cos{(p-q)x}}{p-q}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{px}\cos{qx}dx}\)
\(=\frac{1}{2}\int{2\sin{px}\cos{qx}dx}\)
\(=\frac{1}{2}\int{\{\sin{(px+qx)}+\sin{(px-qx)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int{\sin{(p+q)x}dx}+\frac{1}{2}\int{\sin{(p-q)x}dx}\)
\(=\frac{1}{2}\left\{-\frac{1}{(p+q)}\cos{(p+q)x}\right\}+\frac{1}{2}\left\{-\frac{1}{(p-q)}\cos{(p-q)x}\right\}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cos{(p+q)x}}{2(p+q)}-\frac{\cos{(p-q)x}}{2(p-q)}+c\)
\(=-\frac{1}{2}\left\{\frac{\cos{(p+q)x}}{p+q}+\frac{\cos{(p-q)x}}{p-q}\right\}+c\)
\(=\frac{1}{2}\int{2\sin{px}\cos{qx}dx}\)
\(=\frac{1}{2}\int{\{\sin{(px+qx)}+\sin{(px-qx)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int{\sin{(p+q)x}dx}+\frac{1}{2}\int{\sin{(p-q)x}dx}\)
\(=\frac{1}{2}\left\{-\frac{1}{(p+q)}\cos{(p+q)x}\right\}+\frac{1}{2}\left\{-\frac{1}{(p-q)}\cos{(p-q)x}\right\}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cos{(p+q)x}}{2(p+q)}-\frac{\cos{(p-q)x}}{2(p-q)}+c\)
\(=-\frac{1}{2}\left\{\frac{\cos{(p+q)x}}{p+q}+\frac{\cos{(p-q)x}}{p-q}\right\}+c\)
\(Q.2.(xxvi)\) \(\int{\sin{4x}\sin{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(3\sin{2x}-\sin{6x})+c\)
উত্তরঃ \(\frac{1}{12}(3\sin{2x}-\sin{6x})+c\)
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(6x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 6.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6dx=dt_{2}\)
\(\therefore dx=\frac{1}{6}dt_{2}\)
\(\int{\sin{4x}\sin{2x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(6x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 6.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6dx=dt_{2}\)
\(\therefore dx=\frac{1}{6}dt_{2}\)
\(=\frac{1}{2}\int{2\sin{4x}\sin{2x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(4x-2x)}-\cos{(4x+2x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos{6x}dx}\)
\(=\frac{1}{2}\int{\cos{t_{1}}.\frac{1}{2}dt_{1}}-\frac{1}{2}\int{\cos{t_{2}}.\frac{1}{6}dt_{2}}\)
\(=\frac{1}{4}\int{\cos{t_{1}}dt_{1}}-\frac{1}{12}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{4}\sin{t_{1}}-\frac{1}{12}\sin{t_{2}}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{12}\sin{6x}+c\) ➜ \(\because t_{1}=2x, t_{2}=6x\)
\(=\frac{1}{12}(3\sin{2x}-\sin{6x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin{4x}\sin{2x}dx}\)
\(=\frac{1}{2}\int{2\sin{4x}\sin{2x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(4x-2x)}-\cos{(4x+2x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos{6x}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{2}.\frac{1}{6}\sin{6x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{12}\sin{6x}+c\)
\(=\frac{1}{12}(3\sin{2x}-\sin{6x})+c\)
\(=\frac{1}{2}\int{2\sin{4x}\sin{2x}dx}\)
\(=\frac{1}{2}\int{\{\cos{(4x-2x)}-\cos{(4x+2x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos{6x}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{2}.\frac{1}{6}\sin{6x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{12}\sin{6x}+c\)
\(=\frac{1}{12}(3\sin{2x}-\sin{6x})+c\)
\(Q.2.(xxvii)\) \(\int{3\sin{3x}\cos{4x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{3}{14}(7\cos{x}-\cos{7x})+c\)
উত্তরঃ \(\frac{3}{14}(7\cos{x}-\cos{7x})+c\)
সমাধানঃ
ধরি,
\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(\int{3\sin{3x}\cos{4x}dx}\)\(7x=t\)
\(\Rightarrow \frac{d}{dx}(7x)=\frac{d}{dx}(t)\)
\(\Rightarrow 7.1=\frac{dt}{dx}\)
\(\Rightarrow 7=\frac{dt}{dx}\)
\(\Rightarrow 7dx=dt\)
\(\therefore dx=\frac{1}{7}dt\)
\(=\frac{3}{2}\int{2\cos{4x}\sin{3x}dx}\)
\(=\frac{3}{2}\int{\{\sin{(4x+3x)}-\sin{(4x-3x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=\frac{3}{2}\int{\sin{7x}dx}-\frac{3}{2}\int{\sin{x}dx}\)
\(=\frac{3}{2}\int{\sin{t}.\frac{1}{7}dt}-\frac{3}{2}\int{\sin{x}dx}\)
\(=\frac{3}{14}\int{\sin{t}dt}-\frac{3}{2}\int{\sin{x}dx}\)
\(=\frac{3}{14}(-\cos{t})-\frac{3}{2}(-\cos{x})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{14}\cos{7x}+\frac{3}{2}\cos{x}+c\) ➜ \(\because t=7x\)
\(=\frac{3}{14}(7\cos{x}-\cos{7x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{3\sin{3x}\cos{4x}dx}\)
\(=\frac{3}{2}\int{2\cos{4x}\sin{3x}dx}\)
\(=\frac{3}{2}\int{\{\sin{(4x+3x)}-\sin{(4x-3x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=\frac{3}{2}\int{\sin{7x}dx}-\frac{3}{2}\int{\sin{x}dx}\)
\(=\frac{3}{2}\left\{-\frac{1}{7}\cos{7x}\right\}-\frac{3}{2}(-\cos{x})+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{14}\cos{7x}+\frac{3}{2}\cos{x}+c\)
\(=\frac{3}{14}(7\cos{x}-\cos{7x})+c\)
\(=\frac{3}{2}\int{2\cos{4x}\sin{3x}dx}\)
\(=\frac{3}{2}\int{\{\sin{(4x+3x)}-\sin{(4x-3x)}\}dx}\) ➜ \(\because 2\cos{A}\sin{B}=\sin{(A+B)}-\sin{(A-B)}\)
\(=\frac{3}{2}\int{\sin{7x}dx}-\frac{3}{2}\int{\sin{x}dx}\)
\(=\frac{3}{2}\left\{-\frac{1}{7}\cos{7x}\right\}-\frac{3}{2}(-\cos{x})+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{14}\cos{7x}+\frac{3}{2}\cos{x}+c\)
\(=\frac{3}{14}(7\cos{x}-\cos{7x})+c\)
\(Q.2.(xxviii)\) \(\int{\sin^2{3\theta}d\theta}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(6\theta-\sin{6\theta})+c\)
উত্তরঃ \(\frac{1}{12}(6\theta-\sin{6\theta})+c\)
সমাধানঃ
ধরি,
\(6\theta=t\)
\(\Rightarrow \frac{d}{dx}(6\theta)=\frac{d}{d\theta}(t)\)
\(\Rightarrow 6.1=\frac{dt}{d\theta}\)
\(\Rightarrow 6=\frac{dt}{d\theta}\)
\(\Rightarrow 6d\theta=dt\)
\(\therefore d\theta=\frac{1}{6}dt\)
\(\int{\sin^2{3\theta}d\theta}\)\(6\theta=t\)
\(\Rightarrow \frac{d}{dx}(6\theta)=\frac{d}{d\theta}(t)\)
\(\Rightarrow 6.1=\frac{dt}{d\theta}\)
\(\Rightarrow 6=\frac{dt}{d\theta}\)
\(\Rightarrow 6d\theta=dt\)
\(\therefore d\theta=\frac{1}{6}dt\)
\(=\frac{1}{2}\int{2\sin^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{(1-\cos{6\theta})d\theta}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}-\frac{1}{2}\int{\cos{6\theta}d\theta}\)
\(=\frac{1}{2}\int{d\theta}-\frac{1}{2}\int{\cos{t}\frac{1}{6}dt}\)
\(=\frac{1}{2}\int{d\theta}-\frac{1}{12}\int{\cos{t}dt}\)
\(=\frac{1}{2}\theta-\frac{1}{12}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\theta-\frac{1}{12}\sin{6\theta}+c\) ➜ \(\because t=6\theta\)
\(=\frac{1}{12}(6\theta-\sin{6\theta})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{2\sin^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{(1-\cos{6\theta})d\theta}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}-\frac{1}{2}\int{\cos{6\theta}d\theta}\)
\(=\frac{1}{2}\theta-\frac{1}{2}.\frac{1}{6}\sin{6\theta}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\theta-\frac{1}{12}\sin{6\theta}+c\)
\(=\frac{1}{12}(6\theta-\sin{6\theta})+c\)
\(=\frac{1}{2}\int{2\sin^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{(1-\cos{6\theta})d\theta}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}-\frac{1}{2}\int{\cos{6\theta}d\theta}\)
\(=\frac{1}{2}\theta-\frac{1}{2}.\frac{1}{6}\sin{6\theta}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\theta-\frac{1}{12}\sin{6\theta}+c\)
\(=\frac{1}{12}(6\theta-\sin{6\theta})+c\)
\(Q.2.(xxix)\) \(\int{\cos^2{\frac{\theta}{2}}d\theta}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(\theta+\sin{\theta})+c\)
উত্তরঃ \(\frac{1}{2}(\theta+\sin{\theta})+c\)
সমাধানঃ
\(\int{\cos^2{\frac{\theta}{2}}d\theta}\)
\(=\frac{1}{2}\int{2\cos^2{\frac{\theta}{2}}d\theta}\)
\(=\frac{1}{2}\int{(1+\cos{\theta})d\theta}\) ➜ \(\because 2\cos^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{\theta}d\theta}\)
\(=\frac{1}{2}\theta+\frac{1}{2}\sin{\theta}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}(\theta+\sin{\theta})+c\)
\(=\frac{1}{2}\int{2\cos^2{\frac{\theta}{2}}d\theta}\)
\(=\frac{1}{2}\int{(1+\cos{\theta})d\theta}\) ➜ \(\because 2\cos^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{\theta}d\theta}\)
\(=\frac{1}{2}\theta+\frac{1}{2}\sin{\theta}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}(\theta+\sin{\theta})+c\)
\(Q.2.(xxx)\) \(\int{\cos^2{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{8}(4x+\sin{4x})+c\)
[ ঢঃ২০০০ ]
উত্তরঃ \(\frac{1}{8}(4x+\sin{4x})+c\)
[ ঢঃ২০০০ ]
সমাধানঃ
ধরি,
\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(\int{\cos^2{2x}dx}\)\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(=\frac{1}{2}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{4x}dx}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{t}\frac{1}{4}dt}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{8}\int{\cos{t}dt}\)
\(=\frac{1}{2}x+\frac{1}{8}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}x+\frac{1}{8}\sin{4x}+c\) ➜ \(\because t=4x\)
\(=\frac{1}{8}(4x+\sin{4x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{4x}dx}\)
\(=\frac{1}{2}x+\frac{1}{2}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}x+\frac{1}{8}\sin{4x}+c\)
\(=\frac{1}{8}(4x+\sin{4x})+c\)
\(=\frac{1}{2}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{4x}dx}\)
\(=\frac{1}{2}x+\frac{1}{2}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}x+\frac{1}{8}\sin{4x}+c\)
\(=\frac{1}{8}(4x+\sin{4x})+c\)
\(Q.2.(xxxi)\) \(\int{\cos^2{\frac{x}{2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(x+\sin{x})+c\)
উত্তরঃ \(\frac{1}{2}(x+\sin{x})+c\)
সমাধানঃ
\(\int{\cos^2{\frac{x}{2}}dx}\)
\(=\frac{1}{2}\int{2\cos^2{\frac{x}{2}}dx}\)
\(=\frac{1}{2}\int{(1+\cos{x})dx}\) ➜ \(\because 2\cos^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{x}dx}\)
\(=\frac{1}{2}x+\frac{1}{2}\sin{x}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}(x+\sin{x})+c\)
\(=\frac{1}{2}\int{2\cos^2{\frac{x}{2}}dx}\)
\(=\frac{1}{2}\int{(1+\cos{x})dx}\) ➜ \(\because 2\cos^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{x}dx}\)
\(=\frac{1}{2}x+\frac{1}{2}\sin{x}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}(x+\sin{x})+c\)
\(Q.2.(xxxii)\) \(\int{\sin^3{\theta}d\theta}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(\cos{3\theta}-9\cos{\theta})+c\)
উত্তরঃ \(\frac{1}{12}(\cos{3\theta}-9\cos{\theta})+c\)
সমাধানঃ
ধরি,
\(3\theta=t\)
\(\Rightarrow \frac{d}{d\theta}(3\theta)=\frac{d}{d\theta}(t)\)
\(\Rightarrow 3.1=\frac{dt}{d\theta}\)
\(\Rightarrow 3=\frac{dt}{d\theta}\)
\(\Rightarrow 3d\theta=dt\)
\(\therefore d\theta=\frac{1}{3}dt\)
\(\int{\sin^3{\theta}d\theta}\)\(3\theta=t\)
\(\Rightarrow \frac{d}{d\theta}(3\theta)=\frac{d}{d\theta}(t)\)
\(\Rightarrow 3.1=\frac{dt}{d\theta}\)
\(\Rightarrow 3=\frac{dt}{d\theta}\)
\(\Rightarrow 3d\theta=dt\)
\(\therefore d\theta=\frac{1}{3}dt\)
\(=\frac{1}{4}\int{4\sin^3{\theta}d\theta}\)
\(=\frac{1}{4}\int{(3\sin{\theta}-\sin{3\theta})d\theta}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int{\sin{\theta}d\theta}-\frac{1}{4}\int{\sin{3\theta}d\theta}\)
\(=\frac{3}{4}\int{\sin{\theta}d\theta}-\frac{1}{4}\int{\sin{t}\frac{1}{3}dt}\)
\(=\frac{3}{4}\int{\sin{\theta}d\theta}-\frac{1}{12}\int{\sin{t}dt}\)
\(=\frac{3}{4}(-\cos{\theta})-\frac{1}{12}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{4}\cos{\theta}+\frac{1}{12}\cos{t}+c\)
\(=-\frac{3}{4}\cos{\theta}+\frac{1}{12}\cos{3\theta}+c\) ➜ \(\because t=3\theta\)
\(=\frac{1}{12}(\cos{3\theta}-9\cos{\theta})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin^3{\theta}d\theta}\)
\(=\frac{1}{4}\int{4\sin^3{\theta}d\theta}\)
\(=\frac{1}{4}\int{(3\sin{\theta}-\sin{3\theta})d\theta}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int{\sin{\theta}d\theta}-\frac{1}{4}\int{\sin{3\theta}d\theta}\)
\(=\frac{3}{4}(-\cos{\theta})-\frac{1}{4}\left(-\frac{1}{3}\cos{3\theta}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{4}\cos{\theta}+\frac{1}{12}\cos{3\theta}+c\)
\(=\frac{1}{12}(\cos{3\theta}-9\cos{\theta})+c\)
\(=\frac{1}{4}\int{4\sin^3{\theta}d\theta}\)
\(=\frac{1}{4}\int{(3\sin{\theta}-\sin{3\theta})d\theta}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int{\sin{\theta}d\theta}-\frac{1}{4}\int{\sin{3\theta}d\theta}\)
\(=\frac{3}{4}(-\cos{\theta})-\frac{1}{4}\left(-\frac{1}{3}\cos{3\theta}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{4}\cos{\theta}+\frac{1}{12}\cos{3\theta}+c\)
\(=\frac{1}{12}(\cos{3\theta}-9\cos{\theta})+c\)
\(Q.2.(xxxiii)\) \(\int{\sin^3{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{24}(\cos{6x}-9\cos{2x})+c\)
[ ঢাঃ২০০১]
উত্তরঃ \(\frac{1}{24}(\cos{6x}-9\cos{2x})+c\)
[ ঢাঃ২০০১]
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(6x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 6.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6dx=dt_{2}\)
\(\therefore dx=\frac{1}{6}dt_{2}\)
\(\int{\sin^3{2x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(6x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 6.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6=\frac{dt_{2}}{dx}\)
\(\Rightarrow 6dx=dt_{2}\)
\(\therefore dx=\frac{1}{6}dt_{2}\)
\(=\frac{1}{4}\int{4\sin^3{2x}dx}\)
\(=\frac{1}{4}\int{(3\sin{2x}-\sin{6x})dx}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int{\sin{2x}dx}-\frac{1}{4}\int{\sin{6x}dx}\)
\(=\frac{3}{4}\int{\sin{t_{1}}.\frac{1}{2}dt_{1}}-\frac{1}{4}\int{\sin{t_{2}}.\frac{1}{6}dt_{2}}\)
\(=\frac{3}{8}\int{\sin{t_{1}}dt_{1}}-\frac{1}{24}\int{\sin{t_{2}}dt_{2}}\)
\(=\frac{3}{8}(-\cos{t_{1}})-\frac{1}{24}(-\cos{t_{2}})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{8}\cos{t_{1}}+\frac{1}{24}\cos{t_{2}}+c\)
\(=-\frac{3}{8}\cos{2x}+\frac{1}{24}\cos{6x}+c\) ➜ \(\because t_{1}=2x, t_{2}=6x\)
\(=\frac{1}{24}(\cos{6x}-9\cos{2x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin^3{2x}dx}\)
\(=\frac{1}{4}\int{4\sin^3{2x}dx}\)
\(=\frac{1}{4}\int{(3\sin{2x}-\sin{6x})dx}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int{\sin{2x}dx}-\frac{1}{4}\int{\sin{6x}dx}\)
\(=\frac{3}{4}\left(-\frac{1}{2}\cos{2x}\right)-\frac{1}{4}\left(-\frac{1}{6}\cos{6x}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{8}\cos{2x}+\frac{1}{24}\cos{6x}+c\)
\(=\frac{1}{24}(\cos{6x}-9\cos{2x})+c\)
\(=\frac{1}{4}\int{4\sin^3{2x}dx}\)
\(=\frac{1}{4}\int{(3\sin{2x}-\sin{6x})dx}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int{\sin{2x}dx}-\frac{1}{4}\int{\sin{6x}dx}\)
\(=\frac{3}{4}\left(-\frac{1}{2}\cos{2x}\right)-\frac{1}{4}\left(-\frac{1}{6}\cos{6x}\right)+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{3}{8}\cos{2x}+\frac{1}{24}\cos{6x}+c\)
\(=\frac{1}{24}(\cos{6x}-9\cos{2x})+c\)
\(Q.2.(xxxiv)\) \(\int{\sin^4{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{32}(12x-8\sin{2x}+\sin{4x})+c\)
[ কুঃ২০০৯]
উত্তরঃ \(\frac{1}{32}(12x-8\sin{2x}+\sin{4x})+c\)
[ কুঃ২০০৯]
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(\int{\sin^4{x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(=\frac{1}{4}\int{(2\sin^2{x})^2dx}\)
\(=\frac{1}{4}\int{(1-\cos{2x})^2dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{4}\int{(1-2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{t_{1}}.\frac{1}{2}dt_{1}}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{t_{2}}.\frac{1}{4}dt_{2}}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{4}\int{\cos{t_{1}}dt_{1}}+\frac{1}{8}\int{dx}+\frac{1}{32}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{4}x-\frac{1}{4}\sin{t_{1}}+\frac{1}{8}x+\frac{1}{32}\sin{t_{2}}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x-\frac{1}{4}\sin{t_{1}}+\frac{1}{32}\sin{t_{2}}+c\)
\(=\frac{1}{4}x+\frac{1}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\) ➜ \(\because t_{1}=2x, t_{2}=4x\)
\(=\frac{1+2}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x-8\sin{2x}+\sin{4x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin^4{x}dx}\)
\(=\frac{1}{4}\int{(2\sin^2{x})^2dx}\)
\(=\frac{1}{4}\int{(1-\cos{2x})^2dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{4}\int{(1-2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}x-\frac{1}{2}.\frac{1}{2}\sin{2x}+\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{2+1}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x-8\sin{2x}+\sin{4x})+c\)
\(=\frac{1}{4}\int{(2\sin^2{x})^2dx}\)
\(=\frac{1}{4}\int{(1-\cos{2x})^2dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{4}\int{(1-2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}-\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}x-\frac{1}{2}.\frac{1}{2}\sin{2x}+\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{2+1}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x-\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x-8\sin{2x}+\sin{4x})+c\)
\(Q.2.(xxxv)\) \(\int{\left(1+\cos^2{\frac{x}{2}}\right)dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(3x+\sin{x})+c\)
[ বুটেক্সঃ২০০৪-২০০৫ ]
উত্তরঃ \(\frac{1}{2}(3x+\sin{x})+c\)
[ বুটেক্সঃ২০০৪-২০০৫ ]
সমাধানঃ
\(\int{\left(1+\cos^2{\frac{x}{2}}\right)dx}\)
\(=\int{dx}+\int{\cos^2{\frac{x}{2}}dx}\)
\(=\int{dx}+\frac{1}{2}\int{2\cos^2{\frac{x}{2}}dx}\)
\(=\int{dx}+\frac{1}{2}\int{(1+\cos{x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\int{dx}+\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{x}dx}\)
\(=x+\frac{1}{2}x+\frac{1}{2}\sin{x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2+1}{2}x+\frac{1}{2}\sin{x}+c\)
\(=\frac{3}{2}x+\frac{1}{2}\sin{x}+c\)
\(=\frac{1}{2}(3x+\sin{x})+c\)
\(=\int{dx}+\int{\cos^2{\frac{x}{2}}dx}\)
\(=\int{dx}+\frac{1}{2}\int{2\cos^2{\frac{x}{2}}dx}\)
\(=\int{dx}+\frac{1}{2}\int{(1+\cos{x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\int{dx}+\frac{1}{2}\int{dx}+\frac{1}{2}\int{\cos{x}dx}\)
\(=x+\frac{1}{2}x+\frac{1}{2}\sin{x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2+1}{2}x+\frac{1}{2}\sin{x}+c\)
\(=\frac{3}{2}x+\frac{1}{2}\sin{x}+c\)
\(=\frac{1}{2}(3x+\sin{x})+c\)
\(Q.2.(xxxvi)\) \(\int{\sin^2{x}\cos{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{4x}-4x)+c\)
[ রাঃ২০১২,২০১৩; সিঃ২০১১; কুঃ২০০৭,২০১০,২০১৩,২০১৪; চঃ২০০২,২০০৯; যঃ২০০৫ ]
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-\sin{4x}-4x)+c\)
[ রাঃ২০১২,২০১৩; সিঃ২০১১; কুঃ২০০৭,২০১০,২০১৩,২০১৪; চঃ২০০২,২০০৯; যঃ২০০৫ ]
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(\int{\sin^2{x}\cos{2x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(=\frac{1}{2}\int{2\sin^2{x}\cos{2x}dx}\)
\(=\frac{1}{2}\int{(1-\cos{2x})\cos{2x}dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{(\cos{2x}-\cos^2{2x})dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{dx}-\frac{1}{4}\int{\cos{4x}dx}\)
\(=\frac{1}{2}\int{\cos{t_{1}}.\frac{1}{2}dt_{1}}-\frac{1}{4}\int{dx}-\frac{1}{4}\int{\cos{t_{2}}.\frac{1}{4}dt_{2}}\)
\(=\frac{1}{4}\int{\cos{t_{1}}dt_{1}}-\frac{1}{4}\int{dx}-\frac{1}{16}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{4}\sin{t_{1}}-\frac{1}{4}x-\frac{1}{16}\sin{t_{2}}+c\) ➜ \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{4}x-\frac{1}{16}\sin{4x}+c\) ➜ \(\because t_{1}=2x, t_{2}=4x\)
\(=\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
\(=\frac{1}{16}(4\sin{2x}-\sin{4x}-4x)+c\)
বিকল্প পদ্ধতিঃ
\(\int{\sin^2{x}\cos{2x}dx}\)
\(=\frac{1}{2}\int{2\sin^2{x}\cos{2x}dx}\)
\(=\frac{1}{2}\int{(1-\cos{2x})\cos{2x}dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{(\cos{2x}-\cos^2{2x})dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{dx}-\frac{1}{4}\int{\cos{4x}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{4}x-\frac{1}{4}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{4}x-\frac{1}{16}\sin{4x}+c\)
\(=\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
\(=\frac{1}{16}(4\sin{2x}-\sin{4x}-4x)+c\)
\(=\frac{1}{2}\int{2\sin^2{x}\cos{2x}dx}\)
\(=\frac{1}{2}\int{(1-\cos{2x})\cos{2x}dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{(\cos{2x}-\cos^2{2x})dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{dx}-\frac{1}{4}\int{\cos{4x}dx}\)
\(=\frac{1}{2}.\frac{1}{2}\sin{2x}-\frac{1}{4}x-\frac{1}{4}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{4}x-\frac{1}{16}\sin{4x}+c\)
\(=\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
\(=\frac{1}{16}(4\sin{2x}-\sin{4x}-4x)+c\)
\(Q.2.(xxxvii)\) \(\int{(2\cos{x}+\sin{x})\cos{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
[ ঢাঃ২০০৫ ]
উত্তরঃ \(\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
[ ঢাঃ২০০৫ ]
সমাধানঃ
ধরি,
\(2x=t\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(2x=t\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(\int{(2\cos{x}+\sin{x})\cos{x}dx}\)
\(=\int{(2\cos^2{x}+\sin{x}\cos{x})dx}\)
\(=\int{(2\cos^2{x}+\frac{1}{2}.2\sin{x}\cos{x})dx}\)
\(=\int{(1+\cos{2x}+\frac{1}{2}\sin{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin{A}\cos{A}=\sin{2A}\)
\(=\int{dx}+\int{\cos{2x}dx}+\frac{1}{2}\int{\sin{2x}dx}\)
\(=\int{dx}+\int{\cos{t}.\frac{1}{2}dt}+\frac{1}{2}\int{\sin{t}.\frac{1}{2}dt}\)
\(=\int{dx}+\frac{1}{2}\int{\cos{t}dt}+\frac{1}{4}\int{\sin{t}dt}\)
\(=x+\frac{1}{2}\sin{t}+\frac{1}{4}(-\cos{t})+c\) ➜ \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\), \(\int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\frac{1}{2}\sin{t}-\frac{1}{4}\cos{t}+c\)
\(=x+\frac{1}{2}\sin{2x}-\frac{1}{4}\cos{2x}+c\) ➜ \(\because t=2x\)
\(=\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{(2\cos{x}+\sin{x})\cos{x}dx}\)
\(=\int{(2\cos^2{x}+\sin{x}\cos{x})dx}\)
\(=\int{(2\cos^2{x}+\frac{1}{2}.2\sin{x}\cos{x})dx}\)
\(=\int{(1+\cos{2x}+\frac{1}{2}\sin{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin{A}\cos{A}=\sin{2A}\)
\(=\int{dx}+\int{\cos{2x}dx}+\frac{1}{2}\int{\sin{2x}dx}\)
\(=x+\frac{1}{2}\sin{2x}+\frac{1}{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\), \(\int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\frac{1}{2}\sin{2x}-\frac{1}{4}\cos{2x}+c\)
\(=\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
\(=\int{(2\cos^2{x}+\sin{x}\cos{x})dx}\)
\(=\int{(2\cos^2{x}+\frac{1}{2}.2\sin{x}\cos{x})dx}\)
\(=\int{(1+\cos{2x}+\frac{1}{2}\sin{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin{A}\cos{A}=\sin{2A}\)
\(=\int{dx}+\int{\cos{2x}dx}+\frac{1}{2}\int{\sin{2x}dx}\)
\(=x+\frac{1}{2}\sin{2x}+\frac{1}{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\), \(\int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\frac{1}{2}\sin{2x}-\frac{1}{4}\cos{2x}+c\)
\(=\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
\(Q.2.(xxxviii)\) \(\int{\sin^2{x}\cos^2{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{32}(4x-\sin{4x})+c\)
[ ঢাঃ২০১৩; বঃ২০০৮; চঃ২০০৬; বুয়েটঃ২০০৩-২০০৪ ]
উত্তরঃ \(\frac{1}{32}(4x-\sin{4x})+c\)
[ ঢাঃ২০১৩; বঃ২০০৮; চঃ২০০৬; বুয়েটঃ২০০৩-২০০৪ ]
সমাধানঃ
ধরি,
\(2x=t\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(\int{\sin^2{x}\cos^2{x}dx}\)\(2x=t\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2.1=\frac{dt}{dx}\)
\(\Rightarrow 2=\frac{dt}{dx}\)
\(\Rightarrow 2dx=dt\)
\(\therefore dx=\frac{1}{2}dt\)
\(=\frac{1}{4}\int{4\sin^2{x}\cos^2{x}dx}\)
\(=\frac{1}{4}\int{(2\sin{x}\cos{x})^2dx}\)
\(=\frac{1}{4}\int{(\sin{2x})^2dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(=\frac{1}{4}\int{\sin^2{2x}dx}\)
\(=\frac{1}{8}\int{2\sin^2{2x}dx}\)
\(=\frac{1}{8}\int{(1-\cos{4x})dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{8}\int{dx}-\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{1}{8}\int{\cos{t}.\frac{1}{4}dt}\)
\(=\frac{1}{8}\int{dx}-\frac{1}{32}\int{\cos{t}dt}\)
\(=\frac{1}{8}x-\frac{1}{32}\sin{t}+c\) ➜ \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{8}x-\frac{1}{32}\sin{4x}+c\) ➜ \(\because t=4x\)
\(=\frac{1}{32}(4x-\sin{4x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{(2\cos{x}+\sin{x})\cos{x}dx}\)
\(=\int{(2\cos^2{x}+\sin{x}\cos{x})dx}\)
\(=\int{(2\cos^2{x}+\frac{1}{2}.2\sin{x}\cos{x})dx}\)
\(=\int{(1+\cos{2x}+\frac{1}{2}\sin{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin{A}\cos{A}=\sin{2A}\)
\(=\int{dx}+\int{\cos{2x}dx}+\frac{1}{2}\int{\sin{2x}dx}\)
\(=x+\frac{1}{2}\sin{2x}+\frac{1}{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\), \(\int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\frac{1}{2}\sin{2x}-\frac{1}{4}\cos{2x}+c\)
\(=\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
\(=\int{(2\cos^2{x}+\sin{x}\cos{x})dx}\)
\(=\int{(2\cos^2{x}+\frac{1}{2}.2\sin{x}\cos{x})dx}\)
\(=\int{(1+\cos{2x}+\frac{1}{2}\sin{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin{A}\cos{A}=\sin{2A}\)
\(=\int{dx}+\int{\cos{2x}dx}+\frac{1}{2}\int{\sin{2x}dx}\)
\(=x+\frac{1}{2}\sin{2x}+\frac{1}{2}\left(-\frac{1}{2}\cos{2x}\right)+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\), \(\int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x+\frac{1}{2}\sin{2x}-\frac{1}{4}\cos{2x}+c\)
\(=\frac{1}{4}(4x+2\sin{2x}-\cos{2x})+c\)
\(Q.2.(xxxix)\) \(\int{\cos^2{3\theta}d\theta}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(6\theta+\sin{6\theta})+c\)
উত্তরঃ \(\frac{1}{12}(6\theta+\sin{6\theta})+c\)
সমাধানঃ
ধরি,
\(6\theta=t\)
\(\Rightarrow \frac{d}{dx}(6\theta)=\frac{d}{d\theta}(t)\)
\(\Rightarrow 6.1=\frac{dt}{d\theta}\)
\(\Rightarrow 6=\frac{dt}{d\theta}\)
\(\Rightarrow 6d\theta=dt\)
\(\therefore d\theta=\frac{1}{6}dt\)
\(\int{\cos^2{3\theta}d\theta}\)\(6\theta=t\)
\(\Rightarrow \frac{d}{dx}(6\theta)=\frac{d}{d\theta}(t)\)
\(\Rightarrow 6.1=\frac{dt}{d\theta}\)
\(\Rightarrow 6=\frac{dt}{d\theta}\)
\(\Rightarrow 6d\theta=dt\)
\(\therefore d\theta=\frac{1}{6}dt\)
\(=\frac{1}{2}\int{2\cos^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{(1+\cos{6\theta})d\theta}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{6\theta}d\theta}\)
\(=\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{t}\frac{1}{6}dt}\)
\(=\frac{1}{2}\int{d\theta}+\frac{1}{12}\int{\cos{t}dt}\)
\(=\frac{1}{2}\theta+\frac{1}{12}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\theta+\frac{1}{12}\sin{6\theta}+c\) ➜ \(\because t=6\theta\)
\(=\frac{1}{12}(6\theta+\sin{6\theta})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\cos^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{2\cos^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{(1+\cos{6\theta})d\theta}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{6\theta}d\theta}\)
\(=\frac{1}{2}\theta+\frac{1}{2}.\frac{1}{6}\sin{6\theta}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\theta+\frac{1}{12}\sin{6\theta}+c\)
\(=\frac{1}{12}(6\theta+\sin{6\theta})+c\)
\(=\frac{1}{2}\int{2\cos^2{3\theta}d\theta}\)
\(=\frac{1}{2}\int{(1+\cos{6\theta})d\theta}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int{d\theta}+\frac{1}{2}\int{\cos{6\theta}d\theta}\)
\(=\frac{1}{2}\theta+\frac{1}{2}.\frac{1}{6}\sin{6\theta}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\theta+\frac{1}{12}\sin{6\theta}+c\)
\(=\frac{1}{12}(6\theta+\sin{6\theta})+c\)
\(Q.2.(xL)\) \(\int{\cos^3{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{24}(\sin{6x}+9\sin{2x})+c\)
উত্তরঃ \(\frac{1}{24}(\sin{6x}+9\sin{2x})+c\)
সমাধানঃ
ধরি,
\(6x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 6.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 6=\frac{dt_{1}}{dx}\)
\(\Rightarrow 6dx=dt_{1}\)
\(\therefore dx=\frac{1}{6}dt_{1}\)
আবার,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(\int{\cos^3{2x}dx}\)\(6x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(6x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 6.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 6=\frac{dt_{1}}{dx}\)
\(\Rightarrow 6dx=dt_{1}\)
\(\therefore dx=\frac{1}{6}dt_{1}\)
আবার,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(=\frac{1}{4}\int{4\cos^3{2x}dx}\)
\(=\frac{1}{4}\int{(\cos{6x}+3\cos{2x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int{\cos{6x}dx}+\frac{3}{4}\int{\cos{2x}dx}\)
\(=\frac{1}{4}\int{\cos{t_{1}}.\frac{1}{6}dt_{1}}+\frac{3}{4}\int{\cos{t_{2}}.\frac{1}{2}dt_{2}}\)
\(=\frac{1}{24}\int{\cos{t_{1}}dt_{1}}+\frac{3}{8}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{24}\sin{t_{1}}+\frac{3}{8}\sin{t_{2}}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{24}\sin{6x}+\frac{3}{8}\sin{2x}+c\) ➜ \(\because t_{1}=6x, t_{2}=2x\)
\(=\frac{1}{24}(\sin{6x}+9\sin{2x})+c\)
বিকল্প পদ্ধতিঃ
\(\int{\cos^3{2x}dx}\)
\(=\frac{1}{4}\int{4\cos^3{2x}dx}\)
\(=\frac{1}{4}\int{(\cos{6x}+3\cos{2x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int{\cos{6x}dx}+\frac{3}{4}\int{\cos{2x}dx}\)
\(=\frac{1}{4}.\frac{1}{6}\sin{6x}+\frac{3}{4}.\frac{1}{2}\sin{2x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{24}\sin{6x}+\frac{3}{8}\sin{2x}+c\)
\(=\frac{1}{24}(\sin{6x}+9\sin{2x})+c\)
\(=\frac{1}{4}\int{4\cos^3{2x}dx}\)
\(=\frac{1}{4}\int{(\cos{6x}+3\cos{2x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int{\cos{6x}dx}+\frac{3}{4}\int{\cos{2x}dx}\)
\(=\frac{1}{4}.\frac{1}{6}\sin{6x}+\frac{3}{4}.\frac{1}{2}\sin{2x}+c\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{24}\sin{6x}+\frac{3}{8}\sin{2x}+c\)
\(=\frac{1}{24}(\sin{6x}+9\sin{2x})+c\)
\(Q.2.(xLi)\) \(\int{\sqrt{1+\sin{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2(\sin{\frac{x}{2}}-\cos{\frac{x}{2}})+c\)
উত্তরঃ \(2(\sin{\frac{x}{2}}-\cos{\frac{x}{2}})+c\)
সমাধানঃ
ধরি,
\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(\int{\sqrt{1+\sin{x}}dx}\)\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\therefore dx=2dt\)
\(=\int{\sqrt{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}dx}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\),\(\sin{2A}=2\sin{A}\cos{A}\)
\(=\int{\sqrt{\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2}dx}\) ➜ \(\because a^2+b^2+2ab=(a+b)^2\)
\(=\int{\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)dx}\)
\(=\int{\sin{\frac{x}{2}}dx}+\int{\cos{\frac{x}{2}}dx}\)
\(=\int{\sin{t}.2dt}+\int{\cos{t}.2dt}\)
\(=2\int{\sin{t}dt}+2\int{\cos{t}dt}\)
\(=2(-\cos{t})+2\sin{t}+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-2\cos{t}+2\sin{t}+c\)
\(=2(\sin{t}-\cos{t})+c\)
\(=2\left(\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right)+c\) ➜ \(\because t=\frac{x}{2}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
বিকল্প পদ্ধতিঃ
\(\int{\sqrt{1+\sin{x}}dx}\)
\(=\int{\sqrt{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}dx}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\),\(\sin{2A}=2\sin{A}\cos{A}\)
\(=\int{\sqrt{\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2}dx}\) ➜ \(\because a^2+b^2+2ab=(a+b)^2\)
\(=\int{\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)dx}\)
\(=\int{\sin{\frac{x}{2}}dx}+\int{\cos{\frac{x}{2}}dx}\)
\(=-\frac{1}{\frac{1}{2}}\cos{\frac{x}{2}}+\frac{1}{\frac{1}{2}}\sin{\frac{x}{2}}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-2\cos{\frac{x}{2}}+2\sin{\frac{x}{2}}+c\)
\(=2\left(\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right)+c\)
\(=\int{\sqrt{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}dx}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1\),\(\sin{2A}=2\sin{A}\cos{A}\)
\(=\int{\sqrt{\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2}dx}\) ➜ \(\because a^2+b^2+2ab=(a+b)^2\)
\(=\int{\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)dx}\)
\(=\int{\sin{\frac{x}{2}}dx}+\int{\cos{\frac{x}{2}}dx}\)
\(=-\frac{1}{\frac{1}{2}}\cos{\frac{x}{2}}+\frac{1}{\frac{1}{2}}\sin{\frac{x}{2}}+c\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-2\cos{\frac{x}{2}}+2\sin{\frac{x}{2}}+c\)
\(=2\left(\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right)+c\)
\(Q.2.(xLii)\) \(\int{\frac{1}{1+\sin{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\tan{x}-\sec{x}+c\)
[ চঃ২০১০; যঃ২০১৩]
উত্তরঃ \(\tan{x}-\sec{x}+c\)
[ চঃ২০১০; যঃ২০১৩]
সমাধানঃ
\(\int{\frac{1}{1+\sin{x}}dx}\)
\(=\int{\frac{1-\sin{x}}{(1+\sin{x})(1-\sin{x})}dx}\) ➜ লব ও হরের সহিত \(1-\sin{x}\) গুণ করে।
\(=\int{\frac{1-\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{1-\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{\left(\frac{1}{\cos^2{x}}-\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int{\left(\frac{1}{\cos^2{x}}-\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\)
\(=\int{\left(\sec^2{x}-\sec{x}\tan{x}\right)dx}\)
\(=\int{\sec^2{x}dx}-\int{\sec{x}\tan{x}dx}\)
\(=\tan{x}-\sec{x}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\int{\frac{1-\sin{x}}{(1+\sin{x})(1-\sin{x})}dx}\) ➜ লব ও হরের সহিত \(1-\sin{x}\) গুণ করে।
\(=\int{\frac{1-\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{1-\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{\left(\frac{1}{\cos^2{x}}-\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int{\left(\frac{1}{\cos^2{x}}-\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\)
\(=\int{\left(\sec^2{x}-\sec{x}\tan{x}\right)dx}\)
\(=\int{\sec^2{x}dx}-\int{\sec{x}\tan{x}dx}\)
\(=\tan{x}-\sec{x}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(Q.2.(xLiii)\) \(\int{\frac{1}{1-\cos{3x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{3}\cot{\frac{3x}{2}}+c\)
উত্তরঃ \(-\frac{1}{3}\cot{\frac{3x}{2}}+c\)
সমাধানঃ
ধরি,
\(\frac{3x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{3x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{3}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{3}{2}=\frac{dt}{dx}\)
\(\therefore dx=\frac{2}{3}dt\)
\(\int{\frac{1}{1-\cos{3x}}dx}\)\(\frac{3x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{3x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{3}{2}.1=\frac{dt}{dx}\)
\(\Rightarrow \frac{3}{2}=\frac{dt}{dx}\)
\(\therefore dx=\frac{2}{3}dt\)
\(=\int{\frac{1}{2\sin^2{\frac{3x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(=\frac{1}{2}\int{\frac{1}{\sin^2{\frac{3x}{2}}}dx}\)
\(=\frac{1}{2}\int{cosec^2{\frac{3x}{2}}dx}\)
\(=\frac{1}{2}\int{cosec^2{t}.\frac{2}{3}dt}\)
\(=\frac{1}{2}.\frac{2}{3}\int{cosec^2{t}dt}\)
\(=\frac{1}{3}(-\cot{t})+c\) ➜ \(\because \int{cosec^2{x}dx}=-\cot{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{3}\cot{t}+c\)
\(=-\frac{1}{3}\cot{\frac{3x}{2}}+c\) ➜ \(\because t=\frac{3x}{2}\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1}{1-\cos{3x}}dx}\)
\(=\int{\frac{1}{2\sin^2{\frac{3x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(=\frac{1}{2}\int{\frac{1}{\sin^2{\frac{3x}{2}}}dx}\)
\(=\frac{1}{2}\int{cosec^2{\frac{3x}{2}}dx}\)
\(=-\frac{1}{\frac{3}{2}}\cot{\frac{3x}{2}}+c\) ➜ \(\because \int{cosec \ {ax}dx}=-\frac{1}{a}\cot{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{3}\cot{\frac{3x}{2}}+c\)
\(=\int{\frac{1}{2\sin^2{\frac{3x}{2}}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(=\frac{1}{2}\int{\frac{1}{\sin^2{\frac{3x}{2}}}dx}\)
\(=\frac{1}{2}\int{cosec^2{\frac{3x}{2}}dx}\)
\(=-\frac{1}{\frac{3}{2}}\cot{\frac{3x}{2}}+c\) ➜ \(\because \int{cosec \ {ax}dx}=-\frac{1}{a}\cot{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{3}\cot{\frac{3x}{2}}+c\)
\(Q.2.(xLiv)\) \(\int{\frac{1+e^{5x}}{\sqrt{e^{3x}}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{2}{3}e^{-\frac{3x}{2}}+\frac{2}{7}e^{\frac{7x}{2}}+c\)
উত্তরঃ \(-\frac{2}{3}e^{-\frac{3x}{2}}+\frac{2}{7}e^{\frac{7x}{2}}+c\)
সমাধানঃ
ধরি,
\(-\frac{3x}{2}=t_{1}\)
\(\Rightarrow \frac{d}{dx}\left(-\frac{3x}{2}\right)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow -\frac{3}{2}.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow -\frac{3}{2}=\frac{dt_{1}}{dx}\)
\(\Rightarrow -\frac{3}{2}dx=dt_{1}\)
\(\therefore dx=-\frac{2}{3}dt_{1}\)
আবার,
\(\frac{7x}{2}=t_{2}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{7x}{2}\right)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow \frac{7}{2}.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow \frac{7}{2}=\frac{dt_{2}}{dx}\)
\(\Rightarrow \frac{7}{2}dx=dt_{2}\)
\(\therefore dx=\frac{2}{7}dt_{2}\)
\(\int{\frac{1+e^{5x}}{\sqrt{e^{3x}}}dx}\)\(-\frac{3x}{2}=t_{1}\)
\(\Rightarrow \frac{d}{dx}\left(-\frac{3x}{2}\right)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow -\frac{3}{2}.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow -\frac{3}{2}=\frac{dt_{1}}{dx}\)
\(\Rightarrow -\frac{3}{2}dx=dt_{1}\)
\(\therefore dx=-\frac{2}{3}dt_{1}\)
আবার,
\(\frac{7x}{2}=t_{2}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{7x}{2}\right)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow \frac{7}{2}.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow \frac{7}{2}=\frac{dt_{2}}{dx}\)
\(\Rightarrow \frac{7}{2}dx=dt_{2}\)
\(\therefore dx=\frac{2}{7}dt_{2}\)
\(=\int{\frac{1+e^{5x}}{(e^{3x})^{\frac{1}{2}}}dx}\)
\(=\int{\frac{1+e^{5x}}{e^{\frac{3x}{2}}}dx}\)
\(=\int{\left(\frac{1}{e^{\frac{3x}{2}}}+\frac{e^{5x}}{e^{\frac{3x}{2}}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{5x-\frac{3x}{2}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{\frac{10x-3x}{2}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{\frac{7x}{2}}\right)dx}\)
\(=\int{e^{-\frac{3x}{2}}dx}+\int{e^{\frac{7x}{2}}dx}\)
\(=\int{e^{t_{1}}\left(-\frac{2}{3}dt_{1}\right)}+\int{e^{t_{2}}\frac{2}{7}dt_{2}}\)
\(=-\frac{2}{3}\int{e^{t_{1}}dt_{1}}+\frac{2}{7}\int{e^{t_{2}}dt_{2}}\)
\(=-\frac{2}{3}e^{t_{1}}+\frac{2}{7}e^{t_{2}}+c\) ➜ \(\because \int{e^{x}dx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{3}e^{-\frac{3x}{2}}+\frac{2}{7}e^{\frac{7x}{2}}+c\) ➜ \(\because t_{1}=-\frac{3x}{2}, t_{2}=\frac{7x}{2}\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1+e^{5x}}{\sqrt{e^{3x}}}dx}\)
\(=\int{\frac{1+e^{5x}}{(e^{3x})^{\frac{1}{2}}}dx}\)
\(=\int{\frac{1+e^{5x}}{e^{\frac{3x}{2}}}dx}\)
\(=\int{\left(\frac{1}{e^{\frac{3x}{2}}}+\frac{e^{5x}}{e^{\frac{3x}{2}}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{5x-\frac{3x}{2}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{\frac{10x-3x}{2}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{\frac{7x}{2}}\right)dx}\)
\(=\int{e^{-\frac{3x}{2}}dx}+\int{e^{\frac{7x}{2}}dx}\)
\(=\frac{1}{-\frac{3}{2}}e^{-\frac{3x}{2}}+\frac{1}{\frac{7}{2}}e^{\frac{7x}{2}}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{3}e^{-\frac{3x}{2}}+\frac{2}{7}e^{\frac{7x}{2}}+c\)
\(=\int{\frac{1+e^{5x}}{(e^{3x})^{\frac{1}{2}}}dx}\)
\(=\int{\frac{1+e^{5x}}{e^{\frac{3x}{2}}}dx}\)
\(=\int{\left(\frac{1}{e^{\frac{3x}{2}}}+\frac{e^{5x}}{e^{\frac{3x}{2}}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{5x-\frac{3x}{2}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{\frac{10x-3x}{2}}\right)dx}\)
\(=\int{\left(e^{-\frac{3x}{2}}+e^{\frac{7x}{2}}\right)dx}\)
\(=\int{e^{-\frac{3x}{2}}dx}+\int{e^{\frac{7x}{2}}dx}\)
\(=\frac{1}{-\frac{3}{2}}e^{-\frac{3x}{2}}+\frac{1}{\frac{7}{2}}e^{\frac{7x}{2}}+c\) ➜ \(\because \int{e^{ax}dx}=\frac{1}{a}e^{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{2}{3}e^{-\frac{3x}{2}}+\frac{2}{7}e^{\frac{7x}{2}}+c\)
\(Q.2.(xLv)\) \(\int{\frac{1}{1-\sin{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\tan{x}+\sec{x}+c\)
উত্তরঃ \(\tan{x}+\sec{x}+c\)
সমাধানঃ
\(\int{\frac{1}{1-\sin{x}}dx}\)
\(=\int{\frac{1+\sin{x}}{(1+\sin{x})(1-\sin{x})}dx}\) ➜ লব ও হরের সহিত \(1+\sin{x}\) গুণ করে।
\(=\int{\frac{1+\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{1+\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{\left(\frac{1}{\cos^2{x}}+\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int{\left(\frac{1}{\cos^2{x}}+\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\)
\(=\int{\left(\sec^2{x}+\sec{x}\tan{x}\right)dx}\)
\(=\int{\sec^2{x}dx}+\int{\sec{x}\tan{x}dx}\)
\(=\tan{x}+\sec{x}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\int{\frac{1+\sin{x}}{(1+\sin{x})(1-\sin{x})}dx}\) ➜ লব ও হরের সহিত \(1+\sin{x}\) গুণ করে।
\(=\int{\frac{1+\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{1+\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int{\left(\frac{1}{\cos^2{x}}+\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int{\left(\frac{1}{\cos^2{x}}+\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\)
\(=\int{\left(\sec^2{x}+\sec{x}\tan{x}\right)dx}\)
\(=\int{\sec^2{x}dx}+\int{\sec{x}\tan{x}dx}\)
\(=\tan{x}+\sec{x}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
অনুশীলনী \(10.B / Q.3\)-এর বর্ণনামূলক প্রশ্নসমুহ
যোজিত ফল নির্ণয় করঃ
\(Q.3.(i)\) \(\int{\frac{1}{\sqrt{x+1}-\sqrt{x-1}}dx}\)
উত্তরঃ \(\frac{1}{3}\left\{(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right\}+c\)
[ দিঃ২০১০; রাঃ২০০২; বিআইটিঃ ১৯৯৪-১৯৯৫ ]
\(Q.3.(ii)\) \(\int{\frac{1}{\sqrt{x+2}-\sqrt{x}}dx}\)
উত্তরঃ \(\frac{1}{3}\left\{(x+2)^{\frac{3}{2}}+x^{\frac{3}{2}}\right\}+c\)
[ ঢাঃ২০০০ ]
\(Q.3.(iii)\) \(\int{\frac{dx}{\sqrt{2x+5}+\sqrt{2x-3}}}\)
উত্তরঃ \(\frac{1}{24}\left\{(2x+5)^{\frac{3}{2}}-(2x-3)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(iv)\) \(\int{\frac{dx}{\sqrt{x+1}+\sqrt{x+2}}}\)
উত্তরঃ \(\frac{2}{3}\left\{(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{1}{3}\left\{(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right\}+c\)
[ দিঃ২০১০; রাঃ২০০২; বিআইটিঃ ১৯৯৪-১৯৯৫ ]
\(Q.3.(ii)\) \(\int{\frac{1}{\sqrt{x+2}-\sqrt{x}}dx}\)
উত্তরঃ \(\frac{1}{3}\left\{(x+2)^{\frac{3}{2}}+x^{\frac{3}{2}}\right\}+c\)
[ ঢাঃ২০০০ ]
\(Q.3.(iii)\) \(\int{\frac{dx}{\sqrt{2x+5}+\sqrt{2x-3}}}\)
উত্তরঃ \(\frac{1}{24}\left\{(2x+5)^{\frac{3}{2}}-(2x-3)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(iv)\) \(\int{\frac{dx}{\sqrt{x+1}+\sqrt{x+2}}}\)
উত্তরঃ \(\frac{2}{3}\left\{(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(v)\) \(\int{\frac{5dx}{\sqrt{3x+3}+\sqrt{3x-2}}}\)
উত্তরঃ \(\frac{2}{9}\left\{(3x+3)^{\frac{3}{2}}-(3x-2)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(vi)\) \(\int{\frac{3xdx}{\sqrt{4x+5}-\sqrt{x+5}}}\)
উত্তরঃ \(\frac{1}{6}\left\{(4x+5)^{\frac{3}{2}}+4(x+5)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(vii)\) \(\int{\frac{(x+8)dx}{\sqrt{3x+5}+\sqrt{2x-3}}}\)
উত্তরঃ \(\frac{1}{9}\left\{2(3x+5)^{\frac{3}{2}}-3(2x-3)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{2}{9}\left\{(3x+3)^{\frac{3}{2}}-(3x-2)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(vi)\) \(\int{\frac{3xdx}{\sqrt{4x+5}-\sqrt{x+5}}}\)
উত্তরঃ \(\frac{1}{6}\left\{(4x+5)^{\frac{3}{2}}+4(x+5)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(vii)\) \(\int{\frac{(x+8)dx}{\sqrt{3x+5}+\sqrt{2x-3}}}\)
উত্তরঃ \(\frac{1}{9}\left\{2(3x+5)^{\frac{3}{2}}-3(2x-3)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(i)\) \(\int{\frac{1}{\sqrt{x+1}+\sqrt{x-1}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\left\{(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right\}+c\)
[ দিঃ২০১০; রাঃ২০০২; বিআইটিঃ ১৯৯৪-১৯৯৫ ]
উত্তরঃ \(\frac{1}{3}\left\{(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right\}+c\)
[ দিঃ২০১০; রাঃ২০০২; বিআইটিঃ ১৯৯৪-১৯৯৫ ]
সমাধানঃ
ধরি,
\(x+1=t_{1}\)
\(\Rightarrow \frac{d}{dx}(x+1)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1dx=dt_{1}\)
\(\therefore dx=dt_{1}\)
আবার,
\(x-1=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1dx=dt_{2}\)
\(\therefore dx=dt_{2}\)
\(\int{\frac{1}{\sqrt{x+1}+\sqrt{x-1}}dx}\)\(x+1=t_{1}\)
\(\Rightarrow \frac{d}{dx}(x+1)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1dx=dt_{1}\)
\(\therefore dx=dt_{1}\)
আবার,
\(x-1=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1dx=dt_{2}\)
\(\therefore dx=dt_{2}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1}+\sqrt{x-1})(\sqrt{x+1}-\sqrt{x-1})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x-1})\) গুণ করে।
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1})^2-(\sqrt{x-1})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{x+1-x+1}dx}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{2}dx}\)
\(=\frac{1}{2}\int{(\sqrt{x+1}-\sqrt{x-1})dx}\)
\(=\frac{1}{2}\int{\sqrt{x+1}dx}-\frac{1}{2}\int{\sqrt{x-1}dx}\)
\(=\frac{1}{2}\int{(x+1)^{\frac{1}{2}}dx}-\frac{1}{2}\int{(x-1)^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\int{(t_{1})^{\frac{1}{2}}dt_{1}}-\frac{1}{2}\int{(t_{2})^{\frac{1}{2}}dt_{2}}\)
\(=\frac{1}{2}\frac{(t_{1})^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2}\frac{(t_{2})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(t_{1})^{\frac{1+2}{2}}}{\frac{1+2}{2}}-\frac{1}{2}\frac{(t_{2})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{2}\frac{(t_{1})^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\frac{(t_{2})^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}(t_{1})^{\frac{3}{2}}-\frac{1}{2}.\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(t_{1})^{\frac{3}{2}}-\frac{1}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(x+1)^{\frac{3}{2}}-\frac{1}{3}(x-1)^{\frac{3}{2}}+c\) ➜ \(\because t_{1}=x+1, t_{2}=x-1\)
\(=\frac{1}{3}\left\{(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1}{\sqrt{x+1}+\sqrt{x-1}}dx}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1}+\sqrt{x-1})(\sqrt{x+1}-\sqrt{x-1})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x-1})\) গুণ করে।
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1})^2-(\sqrt{x-1})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{x+1-x+1}dx}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{2}dx}\)
\(=\frac{1}{2}\int{(\sqrt{x+1}-\sqrt{x-1})dx}\)
\(=\frac{1}{2}\int{\sqrt{x+1}dx}-\frac{1}{2}\int{\sqrt{x-1}dx}\)
\(=\frac{1}{2}\int{(x+1)^{\frac{1}{2}}dx}-\frac{1}{2}\int{(x-1)^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\frac{(x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2}\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(x+a)^{n}dx}=\frac{(x+a)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}(x+1)^{\frac{3}{2}}-\frac{1}{2}.\frac{2}{3}(x-1)^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(x+1)^{\frac{3}{2}}-\frac{1}{3}(x-1)^{\frac{3}{2}}+c\)
\(=\frac{1}{3}\left\{(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right\}+c\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1}+\sqrt{x-1})(\sqrt{x+1}-\sqrt{x-1})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x-1})\) গুণ করে।
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1})^2-(\sqrt{x-1})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{x+1-x+1}dx}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x-1}}{2}dx}\)
\(=\frac{1}{2}\int{(\sqrt{x+1}-\sqrt{x-1})dx}\)
\(=\frac{1}{2}\int{\sqrt{x+1}dx}-\frac{1}{2}\int{\sqrt{x-1}dx}\)
\(=\frac{1}{2}\int{(x+1)^{\frac{1}{2}}dx}-\frac{1}{2}\int{(x-1)^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\frac{(x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2}\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(x+a)^{n}dx}=\frac{(x+a)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}(x+1)^{\frac{3}{2}}-\frac{1}{2}.\frac{2}{3}(x-1)^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(x+1)^{\frac{3}{2}}-\frac{1}{3}(x-1)^{\frac{3}{2}}+c\)
\(=\frac{1}{3}\left\{(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(ii)\) \(\int{\frac{1}{\sqrt{x+2}-\sqrt{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\left\{(x+2)^{\frac{3}{2}}+x^{\frac{3}{2}}\right\}+c\)
[ ঢাঃ২০০০ ]
উত্তরঃ \(\frac{1}{3}\left\{(x+2)^{\frac{3}{2}}+x^{\frac{3}{2}}\right\}+c\)
[ ঢাঃ২০০০ ]
সমাধানঃ
ধরি,
\(x+2=t\)
\(\Rightarrow \frac{d}{dx}(x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\Rightarrow 1dx=dt\)
\(\therefore dx=dt\)
\(\int{\frac{1}{\sqrt{x+2}-\sqrt{x}}dx}\)\(x+2=t\)
\(\Rightarrow \frac{d}{dx}(x+2)=\frac{d}{dx}(t)\)
\(\Rightarrow 1+0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\Rightarrow 1dx=dt\)
\(\therefore dx=dt\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{(\sqrt{x+2}+\sqrt{x})(\sqrt{x+2}-\sqrt{x})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+2}+\sqrt{x})\) গুণ করে।
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{(\sqrt{x+2})^2-(\sqrt{x})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{x+2-x}dx}\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{2}dx}\)
\(=\frac{1}{2}\int{(\sqrt{x+2}+\sqrt{x})dx}\)
\(=\frac{1}{2}\int{\sqrt{x+2}dx}+\frac{1}{2}\int{\sqrt{x}dx}\)
\(=\frac{1}{2}\int{(x+2)^{\frac{1}{2}}dx}+\frac{1}{2}\int{x^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\int{t^{\frac{1}{2}}dt}+\frac{1}{2}\int{x^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{2}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}+\frac{1}{2}\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{2}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{2}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}t^{\frac{3}{2}}+\frac{1}{2}.\frac{2}{3}x^{\frac{3}{2}}+c\)
\(=\frac{1}{3}t^{\frac{3}{2}}+\frac{1}{3}x^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(x+2)^{\frac{3}{2}}+\frac{1}{3}x^{\frac{3}{2}}+c\) ➜ \(\because t=x+2\)
\(=\frac{1}{3}\left\{(x+2)^{\frac{3}{2}}+x^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{1}{\sqrt{x+2}-\sqrt{x}}dx}\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{(\sqrt{x+2}+\sqrt{x})(\sqrt{x+2}-\sqrt{x})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+2}+\sqrt{x})\) গুণ করে।
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{(\sqrt{x+2})^2-(\sqrt{x})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{x+2-x}dx}\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{2}dx}\)
\(=\frac{1}{2}\int{(\sqrt{x+2}+\sqrt{x})dx}\)
\(=\frac{1}{2}\int{\sqrt{x+2}dx}+\frac{1}{2}\int{\sqrt{x}dx}\)
\(=\frac{1}{2}\int{(x+2)^{\frac{1}{2}}dx}+\frac{1}{2}\int{x^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\frac{(x+2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{2}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(x+a)^{n}dx}=\frac{(x+a)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(x+2)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{2}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}(x+2)^{\frac{3}{2}}+\frac{1}{2}.\frac{2}{3}x^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(x+2)^{\frac{3}{2}}+\frac{1}{3}x^{\frac{3}{2}}+c\)
\(=\frac{1}{3}\left\{(x+2)^{\frac{3}{2}}+x^{\frac{3}{2}}\right\}+c\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{(\sqrt{x+2}+\sqrt{x})(\sqrt{x+2}-\sqrt{x})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+2}+\sqrt{x})\) গুণ করে।
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{(\sqrt{x+2})^2-(\sqrt{x})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{x+2-x}dx}\)
\(=\int{\frac{\sqrt{x+2}+\sqrt{x}}{2}dx}\)
\(=\frac{1}{2}\int{(\sqrt{x+2}+\sqrt{x})dx}\)
\(=\frac{1}{2}\int{\sqrt{x+2}dx}+\frac{1}{2}\int{\sqrt{x}dx}\)
\(=\frac{1}{2}\int{(x+2)^{\frac{1}{2}}dx}+\frac{1}{2}\int{x^{\frac{1}{2}}dx}\)
\(=\frac{1}{2}\frac{(x+2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{2}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(x+a)^{n}dx}=\frac{(x+a)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{(x+2)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{2}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{2}.\frac{2}{3}(x+2)^{\frac{3}{2}}+\frac{1}{2}.\frac{2}{3}x^{\frac{3}{2}}+c\)
\(=\frac{1}{3}(x+2)^{\frac{3}{2}}+\frac{1}{3}x^{\frac{3}{2}}+c\)
\(=\frac{1}{3}\left\{(x+2)^{\frac{3}{2}}+x^{\frac{3}{2}}\right\}+c\)
\(Q.3.(iii)\) \(\int{\frac{dx}{\sqrt{2x+5}+\sqrt{2x-3}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{24}\left\{(2x+5)^{\frac{3}{2}}-(2x-3)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{1}{24}\left\{(2x+5)^{\frac{3}{2}}-(2x-3)^{\frac{3}{2}}\right\}+c\)
সমাধানঃ
ধরি,
\(2x+5=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x+5)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(2x-3=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x-3)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(\int{\frac{dx}{\sqrt{2x+5}+\sqrt{2x-3}}}\)\(2x+5=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x+5)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার,
\(2x-3=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x-3)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{(\sqrt{2x+5}+\sqrt{2x-3})(\sqrt{2x+5}-\sqrt{2x-3})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{2x+5}-\sqrt{2x-3})\) গুণ করে।
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{(\sqrt{2x+5})^2-(\sqrt{2x-3})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{2x+5-2x+3}dx}\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{8}dx}\)
\(=\frac{1}{8}\int{(\sqrt{2x+5}-\sqrt{2x-3})dx}\)
\(=\frac{1}{8}\int{\sqrt{2x+5}dx}-\frac{1}{8}\int{\sqrt{2x-3}dx}\)
\(=\frac{1}{8}\int{(2x+5)^{\frac{1}{2}}dx}-\frac{1}{8}\int{(2x-3)^{\frac{1}{2}}dx}\)
\(=\frac{1}{8}\int{(t_{1})^{\frac{1}{2}}\frac{1}{2}dt_{1}}-\frac{1}{8}\int{(t_{2})^{\frac{1}{2}}\frac{1}{2}dt_{2}}\)
\(=\frac{1}{16}\int{(t_{1})^{\frac{1}{2}}dt_{1}}-\frac{1}{16}\int{(t_{2})^{\frac{1}{2}}dt_{2}}\)
\(=\frac{1}{16}\frac{(t_{1})^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{16}\frac{(t_{2})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{16}\frac{(t_{1})^{\frac{1+2}{2}}}{\frac{1+2}{2}}-\frac{1}{16}\frac{(t_{2})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{16}\frac{(t_{1})^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{16}\frac{(t_{2})^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{16}.\frac{2}{3}(t_{1})^{\frac{3}{2}}-\frac{1}{16}.\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{1}{24}(t_{1})^{\frac{3}{2}}-\frac{1}{24}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{1}{24}(2x+5)^{\frac{3}{2}}-\frac{1}{24}(2x-3)^{\frac{3}{2}}+c\) ➜ \(\because t_{1}=2x+5, t_{2}=2x-3\)
\(=\frac{1}{24}\left\{(2x+5)^{\frac{3}{2}}-(2x-3)^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{dx}{\sqrt{2x+5}+\sqrt{2x-3}}}\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{(\sqrt{2x+5}+\sqrt{2x-3})(\sqrt{2x+5}-\sqrt{2x-3})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{2x+5}-\sqrt{2x-3})\) গুণ করে।
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{(\sqrt{2x+5})^2-(\sqrt{2x-3})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{2x+5-2x+3}dx}\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{8}dx}\)
\(=\frac{1}{8}\int{(\sqrt{2x+5}-\sqrt{2x-3})dx}\)
\(=\frac{1}{8}\int{\sqrt{2x+5}dx}-\frac{1}{8}\int{\sqrt{2x-3}dx}\)
\(=\frac{1}{8}\int{(2x+5)^{\frac{1}{2}}dx}-\frac{1}{8}\int{(2x-3)^{\frac{1}{2}}dx}\)
\(=\frac{1}{8}.\frac{1}{2}\frac{(2x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{8}.\frac{1}{2}\frac{(2x-3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{16}\frac{(2x+5)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{16}\frac{(2x-3)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{16}.\frac{2}{3}(2x+5)^{\frac{3}{2}}-\frac{1}{16}.\frac{2}{3}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{1}{24}(2x+5)^{\frac{3}{2}}-\frac{1}{24}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{1}{24}\left\{(2x+5)^{\frac{3}{2}}-(2x-3)^{\frac{3}{2}}\right\}+c\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{(\sqrt{2x+5}+\sqrt{2x-3})(\sqrt{2x+5}-\sqrt{2x-3})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{2x+5}-\sqrt{2x-3})\) গুণ করে।
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{(\sqrt{2x+5})^2-(\sqrt{2x-3})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{2x+5-2x+3}dx}\)
\(=\int{\frac{\sqrt{2x+5}-\sqrt{2x-3}}{8}dx}\)
\(=\frac{1}{8}\int{(\sqrt{2x+5}-\sqrt{2x-3})dx}\)
\(=\frac{1}{8}\int{\sqrt{2x+5}dx}-\frac{1}{8}\int{\sqrt{2x-3}dx}\)
\(=\frac{1}{8}\int{(2x+5)^{\frac{1}{2}}dx}-\frac{1}{8}\int{(2x-3)^{\frac{1}{2}}dx}\)
\(=\frac{1}{8}.\frac{1}{2}\frac{(2x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{8}.\frac{1}{2}\frac{(2x-3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{16}\frac{(2x+5)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{16}\frac{(2x-3)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{16}.\frac{2}{3}(2x+5)^{\frac{3}{2}}-\frac{1}{16}.\frac{2}{3}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{1}{24}(2x+5)^{\frac{3}{2}}-\frac{1}{24}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{1}{24}\left\{(2x+5)^{\frac{3}{2}}-(2x-3)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(iv)\) \(\int{\frac{dx}{\sqrt{x+1}+\sqrt{x+2}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\left\{(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{2}{3}\left\{(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right\}+c\)
সমাধানঃ
ধরি,
\(x+1=t_{1}\)
\(\Rightarrow \frac{d}{dx}(x+1)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\therefore dx=dt_{1}\)
আবার,
\(x+2=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x+2)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1+0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\therefore dx=dt_{2}\)
\(\int{\frac{dx}{\sqrt{x+1}+\sqrt{x+2}}}\)\(x+1=t_{1}\)
\(\Rightarrow \frac{d}{dx}(x+1)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 1=\frac{dt_{1}}{dx}\)
\(\therefore dx=dt_{1}\)
আবার,
\(x+2=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x+2)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1+0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\therefore dx=dt_{2}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1}+\sqrt{x+2})(\sqrt{x+1}-\sqrt{x+2})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x+2})\) গুণ করে।
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1})^2-(\sqrt{x+2})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}dx}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{-1}dx}\)
\(=-\int{(\sqrt{x+1}-\sqrt{x+2})dx}\)
\(=-\int{\sqrt{x+1}dx}+\int{\sqrt{x+2}dx}\)
\(=-\int{(x+1)^{\frac{1}{2}}dx}+\int{(x+2)^{\frac{1}{2}}dx}\)
\(=-\int{(t_{1})^{\frac{1}{2}}dt_{1}}+\int{(t_{2})^{\frac{1}{2}}dt_{2}}\)
\(=-\frac{(t_{1})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{(t_{2})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(t_{1})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+\frac{(t_{2})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{(t_{1})^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(t_{2})^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(t_{1})^{\frac{3}{2}}+\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(t_{1})^{\frac{3}{2}}+\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(x+1)^{\frac{3}{2}}+\frac{2}{3}(x+2)^{\frac{3}{2}}+c\) ➜ \(\because t_{1}=x+1, t_{2}=x+2\)
\(=\frac{2}{3}\left\{(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{dx}{\sqrt{x+1}+\sqrt{x+2}}}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1}+\sqrt{x+2})(\sqrt{x+1}-\sqrt{x+2})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x+2})\) গুণ করে।
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1})^2-(\sqrt{x+2})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}dx}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{-1}dx}\)
\(=-\int{(\sqrt{x+1}-\sqrt{x+2})dx}\)
\(=-\int{\sqrt{x+1}dx}+\int{\sqrt{x+2}dx}\)
\(=-\int{(x+1)^{\frac{1}{2}}dx}+\int{(x+2)^{\frac{1}{2}}dx}\)
\(=-\frac{1}{1}\frac{(x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{1}\frac{(x+2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x+2)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(x+1)^{\frac{3}{2}}+\frac{2}{3}(x+2)^{\frac{3}{2}}+c\)
\(=\frac{2}{3}\left\{(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right\}+c\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1}+\sqrt{x+2})(\sqrt{x+1}-\sqrt{x+2})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x+2})\) গুণ করে।
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1})^2-(\sqrt{x+2})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}dx}\)
\(=\int{\frac{\sqrt{x+1}-\sqrt{x+2}}{-1}dx}\)
\(=-\int{(\sqrt{x+1}-\sqrt{x+2})dx}\)
\(=-\int{\sqrt{x+1}dx}+\int{\sqrt{x+2}dx}\)
\(=-\int{(x+1)^{\frac{1}{2}}dx}+\int{(x+2)^{\frac{1}{2}}dx}\)
\(=-\frac{1}{1}\frac{(x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{1}\frac{(x+2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x+2)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(x+1)^{\frac{3}{2}}+\frac{2}{3}(x+2)^{\frac{3}{2}}+c\)
\(=\frac{2}{3}\left\{(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(v)\) \(\int{\frac{5dx}{\sqrt{3x+3}+\sqrt{3x-2}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{9}\left\{(3x+3)^{\frac{3}{2}}-(3x-2)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{2}{9}\left\{(3x+3)^{\frac{3}{2}}-(3x-2)^{\frac{3}{2}}\right\}+c\)
সমাধানঃ
ধরি,
\(3x+3=t_{1}\)
\(\Rightarrow \frac{d}{dx}(3x+3)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 3.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3dx=dt_{1}\)
\(\therefore dx=\frac{1}{3}dt_{1}\)
আবার,
\(3x-2=t_{2}\)
\(\Rightarrow \frac{d}{dx}(3x-2)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 3.1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 3=\frac{dt_{2}}{dx}\)
\(\Rightarrow 3dx=dt_{2}\)
\(\therefore dx=\frac{1}{3}dt_{2}\)
\(\int{\frac{5dx}{\sqrt{3x+3}+\sqrt{3x-2}}}\)\(3x+3=t_{1}\)
\(\Rightarrow \frac{d}{dx}(3x+3)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 3.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3dx=dt_{1}\)
\(\therefore dx=\frac{1}{3}dt_{1}\)
আবার,
\(3x-2=t_{2}\)
\(\Rightarrow \frac{d}{dx}(3x-2)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 3.1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 3=\frac{dt_{2}}{dx}\)
\(\Rightarrow 3dx=dt_{2}\)
\(\therefore dx=\frac{1}{3}dt_{2}\)
\(=5\int{\frac{1}{\sqrt{3x+3}+\sqrt{3x-2}}dx}\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{(\sqrt{3x+3}+\sqrt{3x-2})(\sqrt{3x+3}-\sqrt{3x-2})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{3x+3}-\sqrt{3x-2})\) গুণ করে।
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{(\sqrt{3x+3})^2-(\sqrt{3x-2})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{3x+3-3x+2}dx}\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{5}dx}\)
\(=\int{(\sqrt{3x+3}-\sqrt{3x-2})dx}\)
\(=\int{\sqrt{3x+3}dx}-\int{\sqrt{3x-2}dx}\)
\(=\int{(3x+3)^{\frac{1}{2}}dx}-\int{(3x-2)^{\frac{1}{2}}dx}\)
\(=\int{(t_{1})^{\frac{1}{2}}.\frac{1}{3}dt_{1}}-\int{(t_{2})^{\frac{1}{2}}.\frac{1}{3}dt_{2}}\)
\(=\frac{1}{3}\int{(t_{1})^{\frac{1}{2}}dt_{1}}-\frac{1}{3}\int{(t_{2})^{\frac{1}{2}}dt_{2}}\)
\(=\frac{1}{3}\frac{(t_{1})^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{3}\frac{(t_{2})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\frac{(t_{1})^{\frac{1+2}{2}}}{\frac{1+2}{2}}-\frac{1}{3}\frac{(t_{2})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{3}\frac{(t_{1})^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{3}\frac{(t_{2})^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{3}.\frac{2}{3}(t_{1})^{\frac{3}{2}}-\frac{1}{3}.\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(t_{1})^{\frac{3}{2}}-\frac{2}{9}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(3x+3)^{\frac{3}{2}}-\frac{2}{9}(3x-2)^{\frac{3}{2}}+c\) ➜ \(\because t_{1}=3x+3, t_{2}=3x-2\)
\(=\frac{2}{9}\left\{(3x+3)^{\frac{3}{2}}-(3x-2)^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{5dx}{\sqrt{3x+3}+\sqrt{3x-2}}}\)
\(=5\int{\frac{1}{\sqrt{3x+3}+\sqrt{3x-2}}dx}\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{(\sqrt{3x+3}+\sqrt{3x-2})(\sqrt{3x+3}-\sqrt{3x-2})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{3x+3}-\sqrt{3x-2})\) গুণ করে।
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{(\sqrt{3x+3})^2-(\sqrt{3x-2})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{3x+3-3x+2}dx}\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{5}dx}\)
\(=\int{(\sqrt{3x+3}-\sqrt{3x-2})dx}\)
\(=\int{\sqrt{3x+3}dx}-\int{\sqrt{3x-2}dx}\)
\(=\int{(3x+3)^{\frac{1}{2}}dx}-\int{(3x-2)^{\frac{1}{2}}dx}\)
\(=\frac{1}{3}\frac{(3x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{3}\frac{(3x-2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\frac{(3x+3)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{3}\frac{(3x-2)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{3}.\frac{2}{3}(3x+3)^{\frac{3}{2}}-\frac{1}{3}.\frac{2}{3}(3x-2)^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(3x+3)^{\frac{3}{2}}-\frac{2}{9}(3x-2)^{\frac{3}{2}}+c\)
\(=\frac{2}{9}\left\{(3x+3)^{\frac{3}{2}}-(3x-2)^{\frac{3}{2}}\right\}+c\)
\(=5\int{\frac{1}{\sqrt{3x+3}+\sqrt{3x-2}}dx}\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{(\sqrt{3x+3}+\sqrt{3x-2})(\sqrt{3x+3}-\sqrt{3x-2})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{3x+3}-\sqrt{3x-2})\) গুণ করে।
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{(\sqrt{3x+3})^2-(\sqrt{3x-2})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{3x+3-3x+2}dx}\)
\(=5\int{\frac{\sqrt{3x+3}-\sqrt{3x-2}}{5}dx}\)
\(=\int{(\sqrt{3x+3}-\sqrt{3x-2})dx}\)
\(=\int{\sqrt{3x+3}dx}-\int{\sqrt{3x-2}dx}\)
\(=\int{(3x+3)^{\frac{1}{2}}dx}-\int{(3x-2)^{\frac{1}{2}}dx}\)
\(=\frac{1}{3}\frac{(3x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{3}\frac{(3x-2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\frac{(3x+3)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{3}\frac{(3x-2)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{3}.\frac{2}{3}(3x+3)^{\frac{3}{2}}-\frac{1}{3}.\frac{2}{3}(3x-2)^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(3x+3)^{\frac{3}{2}}-\frac{2}{9}(3x-2)^{\frac{3}{2}}+c\)
\(=\frac{2}{9}\left\{(3x+3)^{\frac{3}{2}}-(3x-2)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(vi)\) \(\int{\frac{3xdx}{\sqrt{4x+5}-\sqrt{x+5}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{6}\left\{(4x+5)^{\frac{3}{2}}+4(x+5)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{1}{6}\left\{(4x+5)^{\frac{3}{2}}+4(x+5)^{\frac{3}{2}}\right\}+c\)
সমাধানঃ
ধরি,
\(4x+5=t_{1}\)
\(\Rightarrow \frac{d}{dx}(4x+5)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 4.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4dx=dt_{1}\)
\(\therefore dx=\frac{1}{4}dt_{1}\)
আবার,
\(x+5=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x+5)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1+0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\therefore dx=dt_{2}\)
\(\int{\frac{3xdx}{\sqrt{4x+5}-\sqrt{x+5}}}\)\(4x+5=t_{1}\)
\(\Rightarrow \frac{d}{dx}(4x+5)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 4.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4dx=dt_{1}\)
\(\therefore dx=\frac{1}{4}dt_{1}\)
আবার,
\(x+5=t_{2}\)
\(\Rightarrow \frac{d}{dx}(x+5)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 1+0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 1=\frac{dt_{2}}{dx}\)
\(\therefore dx=dt_{2}\)
\(=\int{\frac{3x}{\sqrt{4x+5}-\sqrt{x+5}}dx}\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{(\sqrt{4x+5}+\sqrt{x+5})(\sqrt{4x+5}-\sqrt{x+5})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{4x+5}+\sqrt{x+5})\) গুণ করে।
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{(\sqrt{4x+5})^2-(\sqrt{x+5})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{4x+5-x+5}dx}\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5}}{3x}dx}\)
\(=\int{(\sqrt{4x+5}+\sqrt{x+5})dx}\)
\(=\int{\sqrt{4x+5}dx}+\int{\sqrt{x+5}dx}\)
\(=\int{(4x+5)^{\frac{1}{2}}dx}+\int{(x+5)^{\frac{1}{2}}dx}\)
\(=\int{(t_{1})^{\frac{1}{2}}.\frac{1}{4}dt_{1}}+\int{(t_{2})^{\frac{1}{2}}dt_{2}}\)
\(=\frac{1}{4}\int{(t_{1})^{\frac{1}{2}}dt_{1}}+\int{(t_{2})^{\frac{1}{2}}dt_{2}}\)
\(=\frac{1}{4}\frac{(t_{1})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{(t_{2})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\frac{(t_{1})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+\frac{(t_{2})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{4}\frac{(t_{1})^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(t_{2})^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{4}.\frac{2}{3}(t_{1})^{\frac{3}{2}}+\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{1}{6}(t_{1})^{\frac{3}{2}}+\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{1}{6}(4x+5)^{\frac{3}{2}}+\frac{2}{3}(x+5)^{\frac{3}{2}}+c\) ➜ \(\because t_{1}=4x+5, t_{2}=x+5\)
\(=\frac{1}{6}\left\{(4x+5)^{\frac{3}{2}}+4(x+5)^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{3xdx}{\sqrt{4x+5}-\sqrt{x+5}}}\)
\(=\int{\frac{3x}{\sqrt{4x+5}-\sqrt{x+5}}dx}\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{(\sqrt{4x+5}+\sqrt{x+5})(\sqrt{4x+5}-\sqrt{x+5})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{4x+5}+\sqrt{x+5})\) গুণ করে।
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{(\sqrt{4x+5})^2-(\sqrt{x+5})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{4x+5-x+5}dx}\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5}}{3x}dx}\)
\(=\int{(\sqrt{4x+5}+\sqrt{x+5})dx}\)
\(=\int{\sqrt{4x+5}dx}+\int{\sqrt{x+5}dx}\)
\(=\int{(4x+5)^{\frac{1}{2}}dx}+\int{(x+5)^{\frac{1}{2}}dx}\)
\(=\frac{1}{4}\frac{(4x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{1}\frac{(x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\frac{(4x+5)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x+5)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{4}.\frac{2}{3}(4x+5)^{\frac{3}{2}}+\frac{2}{3}(x+5)^{\frac{3}{2}}+c\)
\(=\frac{1}{6}(4x+5)^{\frac{3}{2}}+\frac{2}{3}(x+5)^{\frac{3}{2}}+c\)
\(=\frac{1}{6}\left\{(4x+5)^{\frac{3}{2}}+4(x+5)^{\frac{3}{2}}\right\}+c\)
\(=\int{\frac{3x}{\sqrt{4x+5}-\sqrt{x+5}}dx}\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{(\sqrt{4x+5}+\sqrt{x+5})(\sqrt{4x+5}-\sqrt{x+5})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{4x+5}+\sqrt{x+5})\) গুণ করে।
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{(\sqrt{4x+5})^2-(\sqrt{x+5})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5})}{4x+5-x+5}dx}\)
\(=\int{\frac{3x(\sqrt{4x+5}+\sqrt{x+5}}{3x}dx}\)
\(=\int{(\sqrt{4x+5}+\sqrt{x+5})dx}\)
\(=\int{\sqrt{4x+5}dx}+\int{\sqrt{x+5}dx}\)
\(=\int{(4x+5)^{\frac{1}{2}}dx}+\int{(x+5)^{\frac{1}{2}}dx}\)
\(=\frac{1}{4}\frac{(4x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{1}{1}\frac{(x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\frac{(4x+5)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x+5)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{4}.\frac{2}{3}(4x+5)^{\frac{3}{2}}+\frac{2}{3}(x+5)^{\frac{3}{2}}+c\)
\(=\frac{1}{6}(4x+5)^{\frac{3}{2}}+\frac{2}{3}(x+5)^{\frac{3}{2}}+c\)
\(=\frac{1}{6}\left\{(4x+5)^{\frac{3}{2}}+4(x+5)^{\frac{3}{2}}\right\}+c\)
\(Q.3.(vii)\) \(\int{\frac{(x+8)dx}{\sqrt{3x+5}+\sqrt{2x-3}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{9}\left\{2(3x+5)^{\frac{3}{2}}-3(2x-3)^{\frac{3}{2}}\right\}+c\)
উত্তরঃ \(\frac{1}{9}\left\{2(3x+5)^{\frac{3}{2}}-3(2x-3)^{\frac{3}{2}}\right\}+c\)
সমাধানঃ
ধরি,
\(3x+5=t_{1}\)
\(\Rightarrow \frac{d}{dx}(3x+5)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 3.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3dx=dt_{1}\)
\(\therefore dx=\frac{1}{3}dt_{1}\)
আবার,
\(2x-3=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x-3)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(\int{\frac{(x+8)dx}{\sqrt{2x+5}+\sqrt{2x-3}}}\)\(3x+5=t_{1}\)
\(\Rightarrow \frac{d}{dx}(3x+5)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 3.1+0=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3=\frac{dt_{1}}{dx}\)
\(\Rightarrow 3dx=dt_{1}\)
\(\therefore dx=\frac{1}{3}dt_{1}\)
আবার,
\(2x-3=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x-3)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 2.1-0=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2dx=dt_{2}\)
\(\therefore dx=\frac{1}{2}dt_{2}\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{(\sqrt{3x+5}+\sqrt{2x-3})(\sqrt{3x+5}-\sqrt{2x-3})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{3x+5}-\sqrt{2x-3})\) গুণ করে।
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{(\sqrt{3x+5})^2-(\sqrt{2x-3})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{3x+5-2x+3}dx}\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{x+8}dx}\)
\(=\int{(\sqrt{3x+5}-\sqrt{2x-3})dx}\)
\(=\int{\sqrt{3x+5}dx}-\int{\sqrt{2x-3}dx}\)
\(=\int{(3x+5)^{\frac{1}{2}}dx}-\int{(2x-3)^{\frac{1}{2}}dx}\)
\(=\int{(t_{1})^{\frac{1}{2}}\frac{1}{3}dt_{1}}-\int{(t_{2})^{\frac{1}{2}}\frac{1}{2}dt_{2}}\)
\(=\frac{1}{3}\int{(t_{1})^{\frac{1}{2}}dt_{1}}-\frac{1}{2}\int{(t_{2})^{\frac{1}{2}}dt_{2}}\)
\(=\frac{1}{3}\frac{(t_{1})^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2}\frac{(t_{2})^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\frac{(t_{1})^{\frac{1+2}{2}}}{\frac{1+2}{2}}-\frac{1}{2}\frac{(t_{2})^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{1}{3}\frac{(t_{1})^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\frac{(t_{2})^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{3}.\frac{2}{3}(t_{1})^{\frac{3}{2}}-\frac{1}{2}.\frac{2}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(t_{1})^{\frac{3}{2}}-\frac{1}{3}(t_{2})^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(3x+5)^{\frac{3}{2}}-\frac{1}{3}(2x-3)^{\frac{3}{2}}+c\) ➜ \(\because t_{1}=3x+5, t_{2}=2x-3\)
\(=\frac{1}{9}\left\{2(3x+5)^{\frac{3}{2}}-3(2x-3)^{\frac{3}{2}}\right\}+c\)
বিকল্প পদ্ধতিঃ
\(\int{\frac{(x+8)dx}{\sqrt{2x+5}+\sqrt{2x-3}}}\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{(\sqrt{3x+5}+\sqrt{2x-3})(\sqrt{3x+5}-\sqrt{2x-3})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{3x+5}-\sqrt{2x-3})\) গুণ করে।
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{(\sqrt{3x+5})^2-(\sqrt{2x-3})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{3x+5-2x+3}dx}\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{x+8}dx}\)
\(=\int{(\sqrt{3x+5}-\sqrt{2x-3})dx}\)
\(=\int{\sqrt{3x+5}dx}-\int{\sqrt{2x-3}dx}\)
\(=\int{(3x+5)^{\frac{1}{2}}dx}-\int{(2x-3)^{\frac{1}{2}}dx}\)
\(=\frac{1}{3}\frac{(3x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2}\frac{(2x-3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\frac{(3x+5)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\frac{(2x-3)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{3}.\frac{2}{3}(3x+5)^{\frac{3}{2}}-\frac{1}{2}.\frac{2}{3}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(3x+5)^{\frac{3}{2}}-\frac{1}{3}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{1}{9}\left\{2(3x+5)^{\frac{3}{2}}-3(2x-3)^{\frac{3}{2}}\right\}+c\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{(\sqrt{3x+5}+\sqrt{2x-3})(\sqrt{3x+5}-\sqrt{2x-3})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{3x+5}-\sqrt{2x-3})\) গুণ করে।
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{(\sqrt{3x+5})^2-(\sqrt{2x-3})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{3x+5-2x+3}dx}\)
\(=\int{\frac{(x+8)(\sqrt{3x+5}-\sqrt{2x-3})}{x+8}dx}\)
\(=\int{(\sqrt{3x+5}-\sqrt{2x-3})dx}\)
\(=\int{\sqrt{3x+5}dx}-\int{\sqrt{2x-3}dx}\)
\(=\int{(3x+5)^{\frac{1}{2}}dx}-\int{(2x-3)^{\frac{1}{2}}dx}\)
\(=\frac{1}{3}\frac{(3x+5)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{1}{2}\frac{(2x-3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{(ax+b)^{n}dx}=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\frac{(3x+5)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\frac{(2x-3)^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{1}{3}.\frac{2}{3}(3x+5)^{\frac{3}{2}}-\frac{1}{2}.\frac{2}{3}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{2}{9}(3x+5)^{\frac{3}{2}}-\frac{1}{3}(2x-3)^{\frac{3}{2}}+c\)
\(=\frac{1}{9}\left\{2(3x+5)^{\frac{3}{2}}-3(2x-3)^{\frac{3}{2}}\right\}+c\)
- About
- Author
-
Contact
Call me at +880-1715-651163Email: Golzarrahman1966@gmail.com
Visitors online: 000012