এ অধ্যায়ের পাঠ্যসূচী
- প্রতিস্থাপন পদ্ধতি ( Method of Replacement )
- \(\int{(ax+b)^ndx}\)\(=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\)
- \(\int{e^{ax}dx}\)\(=\frac{1}{a}e^{ax}+c\)
- \(\int{\frac{1}{ax+b}dx}\)\(=\frac{1}{a}\ln{|ax+b|}+c\)
- \(\int{\cos{(ax)}dx}\)\(=\frac{1}{a}\sin{(ax)}+c\)
- \(\int{\sin{(ax)}dx}\)\(=-\frac{1}{a}\cos{(ax)}+c\)
- \(\int{\sec^2{(ax)}dx}\)\(=\frac{1}{a}\tan{(ax)}+c\)
- \(\int{e^{ax+b}dx}\)\(=\frac{1}{a}e^{ax+b}+c\)
- \(\int{\frac{1}{(ax+b)^2}dx}\)\(=-\frac{1}{a}\frac{1}{(ax+b)}+c\)
- \(\int{cosec^2{(ax)}dx}\)\(=-\frac{\cot{(ax)}}{a}+c\)
- \(\int{\sec{(ax)}\tan{(ax)}dx}\)\(=\frac{\sec{(ax)}}{a}+c\)
- \(\int{cosec \ {(ax)}\cot{(ax)}dx}\)\(=-\frac{cosec \ {(ax)}}{a}+c\)
- \(\int{\cos{(ax+b)}dx}\)\(=\frac{\sin{(ax+b)}}{a}+c\)
- \(\int{\sin{(ax+b)}dx}\)\(=-\frac{\cos{(ax+b)}}{a}+c\)
- \(\int{\sec^2{(ax+b)}dx}\)\(=\frac{1}{a}\tan{(ax+b)}+c\)
- \(\int{a^{mx+n}dx}\)\(=\frac{a^{mx+n}}{m\ln{a}}+c\)
- অধ্যায় \(x.B\)-এর উদাহরণসমুহ
- অধ্যায় \(x.B\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
- অধ্যায় \(x.B\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(x.B\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(x.B\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
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।
\(\int{f(ax+b)dx}\)
\(=\int{f(t).\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{f(t)dt}\)
\(\therefore \int{f(ax+b)dx}=\frac{1}{a}\int{f(t)dt}\) এর পর প্রমিত ফাংশনের সূত্র প্রয়োগ করে যোগজীকরণ করতে হয়।
\(\int{f(ax+b)dx}\) এর ক্ষেত্রে \(ax+b\) কে \(t\) ধরতে হয়।
\(\int{f(ax+b)dx}\)
\(=\int{f(t).\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{f(t)dt}\)
\(\therefore \int{f(ax+b)dx}=\frac{1}{a}\int{f(t)dt}\) এর পর প্রমিত ফাংশনের সূত্র প্রয়োগ করে যোগজীকরণ করতে হয়।
\((ax+b)^n\) এবং \(e^{ax}\) এর যোগজীকরণ
Interpretation of \((ax+b)^n\) and \(e^{ax}\)
যোগজীকরণের সূত্র
\(\int{(ax+b)^ndx}\)\(=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\)
\(\int{(ax+b)^ndx}\)\(=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\)
যোগজীকরণের সূত্র
\(\int{e^{ax}dx}\)\(=\frac{1}{a}e^{ax}+c\)
\(\int{e^{ax}dx}\)\(=\frac{1}{a}e^{ax}+c\)
\(\frac{1}{ax+b}\) এবং \(\cos{(ax)}\) এর যোগজীকরণ
Interpretation of \(\frac{1}{ax+b}\) and \(\cos{(ax)}\)
যোগজীকরণের সূত্র
\(\int{\frac{1}{ax+b}dx}\)\(=\frac{1}{a}\ln{|ax+b|}+c\)
\(\int{\frac{1}{ax+b}dx}\)\(=\frac{1}{a}\ln{|ax+b|}+c\)
যোগজীকরণের সূত্র
\(\int{\cos{(ax)}dx}\)\(=\frac{1}{a}\sin{(ax)}+c\)
\(\int{\cos{(ax)}dx}\)\(=\frac{1}{a}\sin{(ax)}+c\)
\(\sin{(ax)}\) এবং \(\sec^2{(ax)}\) এর যোগজীকরণ
Interpretation of \(\sin{(ax)}\) and \(\sec^2{(ax)}\)
যোগজীকরণের সূত্র
\(\int{\sin{(ax)}dx}\)\(=-\frac{1}{a}\cos{(ax)}+c\)
\(\int{\sin{(ax)}dx}\)\(=-\frac{1}{a}\cos{(ax)}+c\)
যোগজীকরণের সূত্র
\(\int{\sec^2{(ax)}dx}\)\(=\frac{1}{a}\tan{(ax)}+c\)
\(\int{\sec^2{(ax)}dx}\)\(=\frac{1}{a}\tan{(ax)}+c\)
\(e^{ax+b}\) এবং \(\frac{1}{(ax+b)^2}\) এর যোগজীকরণ
Interpretation of \(e^{ax+b}\) and \(\frac{1}{(ax+b)^2}\)
যোগজীকরণের সূত্র
\(\int{e^{ax+b}dx}\)\(=\frac{1}{a}e^{ax+b}+c\)
\(\int{e^{ax+b}dx}\)\(=\frac{1}{a}e^{ax+b}+c\)
যোগজীকরণের সূত্র
\(\int{\frac{1}{(ax+b)^2}dx}\)\(=-\frac{1}{a}\frac{1}{(ax+b)}+c\)
\(\int{\frac{1}{(ax+b)^2}dx}\)\(=-\frac{1}{a}\frac{1}{(ax+b)}+c\)
\(cosec^2{(ax)}\) এবং \(\sec{(ax)}\tan{(ax)}\) এর যোগজীকরণ
Interpretation of \(cosec^2{(ax)}\) and \(\sec{(ax)}\tan{(ax)}\)
যোগজীকরণের সূত্র
\(\int{cosec^2{(ax)}dx}\)\(=-\frac{\cot{(ax)}}{a}+c\)
\(\int{cosec^2{(ax)}dx}\)\(=-\frac{\cot{(ax)}}{a}+c\)
যোগজীকরণের সূত্র
\(\int{\sec{(ax)}\tan{(ax)}dx}\)\(=\frac{\sec{(ax)}}{a}+c\)
\(\int{\sec{(ax)}\tan{(ax)}dx}\)\(=\frac{\sec{(ax)}}{a}+c\)
\(cosec \ {(ax)}\cot{(ax)}\) এবং \(\cos{(ax+b)}\) এর যোগজীকরণ
Interpretation of \(cosec \ {(ax)}\cot{(ax)}\) and \(\cos{(ax+b)}\)
যোগজীকরণের সূত্র
\(\int{cosec \ {(ax)}\cot{(ax)}dx}\)\(=-\frac{cosec \ {(ax)}}{a}+c\)
\(\int{cosec \ {(ax)}\cot{(ax)}dx}\)\(=-\frac{cosec \ {(ax)}}{a}+c\)
যোগজীকরণের সূত্র
\(\int{\cos{(ax+b)}dx}\)\(=\frac{\sin{(ax+b)}}{a}+c\)
\(\int{\cos{(ax+b)}dx}\)\(=\frac{\sin{(ax+b)}}{a}+c\)
\(\sin{(ax+b)}\) এবং \(\sec^2{(ax+b)}\) এর যোগজীকরণ
Interpretation of \(\sin{(ax+b)}\) and \(\sec^2{(ax+b)}\)
যোগজীকরণের সূত্র
\(\int{\sin{(ax+b)}dx}\)\(=-\frac{\cos{(ax+b)}}{a}+c\)
\(\int{\sin{(ax+b)}dx}\)\(=-\frac{\cos{(ax+b)}}{a}+c\)
যোগজীকরণের সূত্র
\(\int{\sec^2{(ax+b)}dx}\)\(=\frac{1}{a}\tan{(ax+b)}+c\)
\(\int{\sec^2{(ax+b)}dx}\)\(=\frac{1}{a}\tan{(ax+b)}+c\)
\(a^{mx+n}\) এর যোগজীকরণ
Interpretation of \(a^{mx+n}\)
যোগজীকরণের সূত্র
\(\int{a^{mx+n}dx}\)\(=\frac{a^{mx+n}}{m\ln{a}}+c\)
\(\int{a^{mx+n}dx}\)\(=\frac{a^{mx+n}}{m\ln{a}}+c\)
×
প্রমাণ কর যে, \(\int{(ax+b)^ndx}=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\)
\(=\int{z^n.\frac{1}{a}dz}\)
\(=\frac{1}{a}.\int{z^ndz}\)
\(=\frac{1}{a}.\frac{z^{n+1}}{n+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{(ax+b)^ndx}\)\(=\int{z^n.\frac{1}{a}dz}\)
\(=\frac{1}{a}.\int{z^ndz}\)
\(=\frac{1}{a}.\frac{z^{n+1}}{n+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{e^{ax}dx}=\frac{1}{a}e^{ax}+c\)
\(=\int{e^z.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{e^zdz}\)
\(=\frac{1}{a}e^z+c\) ➜ \(\because \int{e^xdx}=e^x+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{e^{ax}dx}\)\(=\int{e^z.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{e^zdz}\)
\(=\frac{1}{a}e^z+c\) ➜ \(\because \int{e^xdx}=e^x+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\frac{1}{ax+b}dx}=\frac{1}{a}\ln{|ax+b|}+c \)
\(=\int{\frac{1}{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\frac{1}{z}dz}\)
\(=\frac{1}{a}\ln{|z|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|ax+b|}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\frac{1}{ax+b}dx}\)\(=\int{\frac{1}{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\frac{1}{z}dz}\)
\(=\frac{1}{a}\ln{|z|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|ax+b|}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\cos{ax}dx}=\frac{1}{a}\sin{ax}+c\)
\(=\int{\cos{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\cos{z}dz}\)
\(=\frac{1}{a}\sin{z}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\sin{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\cos{ax}dx}\)\(=\int{\cos{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\cos{z}dz}\)
\(=\frac{1}{a}\sin{z}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\sin{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}+c\)
\(=\int{\sin{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sin{z}dz}\)
\(=\frac{1}{a}(-\cos{z})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{a}\cos{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\sin{ax}dx}\)\(=\int{\sin{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sin{z}dz}\)
\(=\frac{1}{a}(-\cos{z})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{a}\cos{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}+c\)
\(=\int{\sec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec^2{z}dz}\)
\(=\frac{1}{a}\tan{z}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\tan{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\sec^2{ax}dx}\)\(=\int{\sec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec^2{z}dz}\)
\(=\frac{1}{a}\tan{z}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\tan{ax}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{e^{ax+b}dx}=\frac{1}{a}e^{ax+b}+c\)
\(=\int{e^{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{e^{z}dz}\)
\(=\frac{1}{a}e^{z}+c\) ➜ \(\because \int{e^{x}dx}=e^{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{ax+b}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{e^{ax+b}dx}\)\(=\int{e^{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{e^{z}dz}\)
\(=\frac{1}{a}e^{z}+c\) ➜ \(\because \int{e^{x}dx}=e^{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{ax+b}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}\frac{1}{(ax+b)}+c\)
\(=\int{\frac{1}{z^2}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\frac{1}{z^2}dz}\)
\(=\frac{1}{a}\int{z^{-2}dz}\)
\(=\frac{1}{a}\frac{z^{-2+1}}{-2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}+c\) 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)
Proof:
\(L.H=\int{\frac{1}{(ax+b)^2}dx}\)\(=\int{\frac{1}{z^2}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\frac{1}{z^2}dz}\)
\(=\frac{1}{a}\int{z^{-2}dz}\)
\(=\frac{1}{a}\frac{z^{-2+1}}{-2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}+c\) 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)
×
প্রমাণ কর যে, \(\int{cosec^2{ax}dx}=-\frac{\cot{ax}}{a}+c\)
\(=\int{cosec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{cosec^2{z}dz}\)
\(=-\frac{1}{a}\cot{z}+c\) ➜ \(\because \int{cosec^2{x}dx}=-\cot{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cot{ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{cosec^2{ax}dx}\)\(=\int{cosec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{cosec^2{z}dz}\)
\(=-\frac{1}{a}\cot{z}+c\) ➜ \(\because \int{cosec^2{x}dx}=-\cot{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cot{ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\sec{ax}\tan{ax}dx}=\frac{\sec{ax}}{a}+c\)
\(=\int{\sec{z}\tan{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec{z}\tan{z}dz}\)
\(=\frac{1}{a}\sec{z}+c\) ➜ \(\because \int{\sec{x}\tan{x}dx}=\sec{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\sec{ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\sec{ax}\tan{ax}dx}\)\(=\int{\sec{z}\tan{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec{z}\tan{z}dz}\)
\(=\frac{1}{a}\sec{z}+c\) ➜ \(\because \int{\sec{x}\tan{x}dx}=\sec{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\sec{ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{cosec \ {ax}\cot{ax}dx}=-\frac{cosec \ {ax}}{a}+c\)
\(=\int{cosec \ {z}\cot{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{cosec \ {z}\cot{z}dz}\)
\(=\frac{1}{a}(-cosec \ {z})+c\) ➜ \(\because \int{cosec \ {x}\cot{x}dx}=-cosec \ {x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{cosec \ {ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{cosec \ {ax}\cot{ax}dx}\)\(=\int{cosec \ {z}\cot{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{cosec \ {z}\cot{z}dz}\)
\(=\frac{1}{a}(-cosec \ {z})+c\) ➜ \(\because \int{cosec \ {x}\cot{x}dx}=-cosec \ {x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{cosec \ {ax}}{a}+c\) ➜ \(\because z=ax\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\cos{(ax+b)}dx}=\frac{\sin{(ax+b)}}{a}+c\)
\(=\int{\cos{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\cos{z}dz}\)
\(=\frac{1}{a}\sin{z}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\sin{(ax+b)}}{a}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\cos{(ax+b)}dx}\)\(=\int{\cos{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\cos{z}dz}\)
\(=\frac{1}{a}\sin{z}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\sin{(ax+b)}}{a}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\sin{(ax+b)}dx}=-\frac{\cos{(ax+b)}}{a}+c\)
\(=\int{\sin{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sin{z}dz}\)
\(=\frac{1}{a}(-\cos{z})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cos{(ax+b)}}{a}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\sin{(ax+b)}dx}\)\(=\int{\sin{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sin{z}dz}\)
\(=\frac{1}{a}(-\cos{z})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{\cos{(ax+b)}}{a}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{\sec^2{(ax+b)}dx}=\frac{1}{a}\tan{(ax+b)}+c\)
\(=\int{\sec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec^2{z}dz}\)
\(=\frac{1}{a}\tan{z}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\tan{(ax+b)}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
Proof:
\(L.H=\int{\sec^2{(ax+b)}dx}\)\(=\int{\sec^2{z}.\frac{1}{a}dz}\)
\(=\frac{1}{a}\int{\sec^2{z}dz}\)
\(=\frac{1}{a}\tan{z}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}+c\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\tan{(ax+b)}+c\) ➜ \(\because z=ax+b\)
\(=R.H\)
(Proved)
×
প্রমাণ কর যে, \(\int{a^{mx+n}dx}=\frac{a^{mx+n}}{m\ln{a}}+c\)
\(=\int{a^{z}.\frac{1}{m}dz}\)
\(=\frac{1}{m}\int{a^{z}dz}\)
\(=\frac{1}{m}\frac{a^{z}}{\ln{a}}+c\) ➜ \(\because \int{a^xdx}=\frac{a^x}{\ln{a}}+c\) 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)
Proof:
\(L.H=\int{a^{mx+n}dx}\)\(=\int{a^{z}.\frac{1}{m}dz}\)
\(=\frac{1}{m}\int{a^{z}dz}\)
\(=\frac{1}{m}\frac{a^{z}}{\ln{a}}+c\) ➜ \(\because \int{a^xdx}=\frac{a^x}{\ln{a}}+c\) 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)
- About
- Author
-
Contact
Call me at +880-1715-651163Email: Golzarrahman1966@gmail.com
Visitors online: 000004