\(L.H=\int{f\{g(x)\}g^{\prime}(x)dx}\)
\(=\int{f(t)dt}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{\{f(x)\}^{n}f^{\prime}(x)dx}\)
\(=\int{t^{n}dt}\)
\(=\frac{t^{n+1}}{n+1}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\{f(x)\}^{n+1}}{n+1}+c\) ➜Note \(\because t=f(x)\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{e^{f(x)}f^{\prime}(x)dx}\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜Note \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{f(x)}+c\) ➜Note \(\because t=f(x)\)
\(=R.H\) \(L.H=R.H\) (Proved)

\(L.H=\int{\frac{f^{\prime}(x)}{\sqrt{f(x)}}dx}\)
\(=\int{\frac{1}{\sqrt{f(x)}}f^{\prime}(x)dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{f(x)}+c\) ➜Note \(\because t=f(x)\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{\frac{f^{\prime}(x)}{f(x)}dx}\)
\(=\int{\frac{1}{f(x)}f^{\prime}(x)dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|f(x)|}+c\) ➜Note \(\because t=f(x)\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{\tan{x}dx}\)
\(=\int{\frac{\sin{x}}{\cos{x}}dx}\)
\(=\int{\frac{1}{\cos{x}}\sin{x}dx}\)
\(=\int{\frac{1}{t}(-dt)}\)
\(=-\int{\frac{1}{t}dt}\)
\(=-\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\ln{|\cos{x}|}+c\) ➜Note \(\because t=\cos{x}\)
\(=\ln{|(\cos{x})^{-1}|}+c\) ➜Note \(\because n\ln{|x|}=\ln{x^n}\)
\(=\ln{|\frac{1}{\cos{x}}|}+c\)
\(=\ln{|\sec{x}|}+c\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{\cot{x}dx}\)
\(=\int{\frac{\cos{x}}{\sin{x}}dx}\)
\(=\int{\frac{1}{\sin{x}}\cos{x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sin{x}|}+c\) ➜Note \(\because t=\sin{x}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{\sec{x}dx}\)
\(=\int{\frac{\sec{x}(\sec{x}+\tan{x})}{(\sec{x}+\tan{x})}dx}\)
\(=\int{\frac{1}{(\sec{x}+\tan{x})}\sec{x}(\sec{x}+\tan{x})dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sec{x}+\tan{x}|}+c\) ➜Note \(\because t=\sec{x}+\tan{x}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\ln{|\sec{x}+\tan{x}|}+c\)
\(=\ln{\left|\frac{1}{\cos{x}}+\frac{\sin{x}}{\cos{x}}\right|}+c\)
\(=\ln{\left|\frac{1+\sin{x}}{\cos{x}}\right|}+c\)
\(=\ln{\left|\frac{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}\right|}+c\) ➜Note \(\because \sin^2{\frac{A}{2}}+\cos^2{\frac{A}{2}}=1, 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}=\sin{A}\), \(\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}=\cos{A}\)
\(=\ln{\left|\frac{(\sin{\frac{x}{2}}+\cos{\frac{x}{2}})^2}{(\cos{\frac{x}{2}}+\sin{\frac{x}{2}})(\cos{\frac{x}{2}}-\sin{\frac{x}{2}})}\right|}+c\) ➜Note \(\because a^2+b^2+2ab=(a+b)^2, (a+b)(a-b)=a^2-b^2\)
\(=\ln{\left|\frac{\sin{\frac{x}{2}}+\cos{\frac{x}{2}}}{\cos{\frac{x}{2}}-\sin{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\frac{\cos{\frac{x}{2}}+\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}-\sin{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\frac{\cos{\frac{x}{2}}\left(1+\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right)}{\cos{\frac{x}{2}}\left(1-\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right)}\right|}+c\)
\(=\ln{\left|\frac{1+\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}}{1-\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}}\right|}+c\)
\(=\ln{\left|\frac{1+\tan{\frac{x}{2}}}{1-\tan{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\frac{\tan{\frac{\pi}{4}}+\tan{\frac{x}{2}}}{1-\tan{\frac{\pi}{4}}\tan{\frac{x}{2}}}\right|}+c\) ➜Note \(\because \tan{\frac{\pi}{4}}=1\)
\(=\ln{\left|\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}\right|}+c\) ➜Note \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\therefore \int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}+c=\ln{\left|\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}\right|}+c\)
(proved)

\(L.H=\int{cosec \ {x}dx}\)
\(=\int{\frac{cosec \ {x}(cosec \ {x}-\cot{x})}{(cosec \ {x}-\cot{x})}dx}\)
\(=\int{\frac{1}{(cosec \ {x}-\cot{x})}cosec \ {x}(cosec \ {x}-\cot{x})dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|cosec \ {x}-\cot{x}|}+c\) ➜Note \(\because t=cosec \ {x}-\cot{x}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\ln{|cosec \ {x}-\cot{x}|}+c\)
\(=\ln{\left|\frac{1}{\sin{x}}-\frac{\cos{x}}{\sin{x}}\right|}+c\)
\(=\ln{\left|\frac{1-\cos{x}}{\sin{x}}\right|}+c\)
\(=\ln{\left|\frac{2\sin^2{\frac{x}{2}}}{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\right|}+c\) ➜Note \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(=\ln{\left|\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\tan{\frac{x}{2}}\right|}+c\)
\(\therefore \int{cosec \ {x}dx}=\ln{|cosec \ {x}-\cot{x}|}+c=\ln{\left|\tan{\frac{x}{2}}\right|}+c\)
(proved)

\(L.H=\int{\tan{(ax+b)}dx}\)
\(=\int{\frac{\sin{(ax+b)}}{\cos{(ax+b)}}dx}\)
\(=\int{\frac{1}{\cos{(ax+b)}}\sin{(ax+b)}dx}\)
\(=\int{\frac{1}{t}\times{-\frac{1}{a}dt}}\)
\(=-\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=-\frac{1}{a}\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{a}\ln{|\cos{(ax+b)}|}+c\) ➜Note \(\because t=\cos{(ax+b)}\)
\(=\frac{1}{a}\ln{|(\cos{(ax+b)})^{-1}|}+c\) ➜Note \(\because n\ln{|x|}=\ln{x^n}\)
\(=\frac{1}{a}\ln{|\frac{1}{\cos{(ax+b)}}|}+c\)
\(=\frac{1}{a}\ln{|\sec{(ax+b)}|}+c\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{\cot{(ax+b)}dx}\)
\(=\int{\frac{\cos{(ax+b)}}{\sin{(ax+b)}}dx}\)
\(=\int{\frac{1}{\sin{(ax+b)}}\cos{(ax+b)}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=\frac{1}{a}\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|\sin{(ax+b)}|}+c\) ➜Note \(\because t=\sin{(ax+b)}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)

\(L.H=\int{\sec{(ax+b)}dx}\)
\(=\int{\frac{\sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}}{\sec{(ax+b)}+\tan{(ax+b)}}dx}\)
\(=\int{\frac{1}{\sec{(ax+b)}+\tan{(ax+b)}}\sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=\frac{1}{a}\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c\) ➜Note \(\because t=\sec{(ax+b)}+\tan{(ax+b)}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1}{\cos{(ax+b)}}+\frac{\sin{(ax+b)}}{\cos{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1+\sin{(ax+b)}}{\cos{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\sin^2{\frac{(ax+b)}{2}}+\cos^2{\frac{(ax+b)}{2}}+2\sin{\frac{(ax+b)}{2}}\cos{\frac{(ax+b)}{2}}}{\cos^2{\frac{(ax+b)}{2}}-\sin^2{\frac{(ax+b)}{2}}}\right|}+c\) ➜Note \(\because \sin^2{\frac{A}{2}}+\cos^2{\frac{A}{2}}=1, 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}=\sin{A}\), \(\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}=\cos{A}\)
\(=\frac{1}{a}\ln{\left|\frac{(\sin{\frac{(ax+b)}{2}}+\cos{\frac{(ax+b)}{2}})^2}{(\cos{\frac{(ax+b)}{2}}+\sin{\frac{(ax+b)}{2}})(\cos{\frac{(ax+b)}{2}}-\sin{\frac{(ax+b)}{2}})}\right|}+c\) ➜Note \(\because a^2+b^2+2ab=(a+b)^2, (a+b)(a-b)=a^2-b^2\)
\(=\frac{1}{a}\ln{\left|\frac{\sin{\frac{(ax+b)}{2}}+\cos{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}-\sin{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\cos{\frac{(ax+b)}{2}}+\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}-\sin{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\cos{\frac{(ax+b)}{2}}\left(1+\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}\right)}{\cos{\frac{(ax+b)}{2}}\left(1-\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}\right)}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1+\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}}{1-\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1+\tan{\frac{(ax+b)}{2}}}{1-\tan{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\tan{\frac{\pi}{4}}+\tan{\frac{(ax+b)}{2}}}{1-\tan{\frac{\pi}{4}}\tan{\frac{(ax+b)}{2}}}\right|}+c\) ➜Note \(\because \tan{\frac{\pi}{4}}=1\)
\(=\frac{1}{a}\ln{\left|\tan{\left\{\frac{\pi}{4}+\frac{(ax+b)}{2}\right\}}\right|}+c\) ➜Note \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\therefore \int{\sec{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c=\frac{1}{a}\ln{\left|\tan{\left\{\frac{\pi}{4}+\frac{(ax+b)}{2}\right\}}\right|}+c\)
(proved)

\(L.H=\int{cosec \ {(ax+b)}dx}\)
\(=\int{\frac{cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}}{cosec \ {(ax+b)}-\cot{(ax+b)}}dx}\)
\(=\int{\frac{1}{cosec \ {(ax+b)}-\cot{(ax+b)}}cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=\frac{1}{a}\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c\) ➜Note \(\because t=cosec \ {(ax+b)}-\cot{(ax+b)}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1}{\sin{(ax+b)}}-\frac{\cos{(ax+b)}}{\sin{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1-\cos{(ax+b)}}{\sin{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{2\sin^2{\frac{(ax+b)}{2}}}{2\sin{\frac{(ax+b)}{2}}\cos{\frac{(ax+b)}{2}}}\right|}+c\) ➜Note \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(=\frac{1}{a}\ln{\left|\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\tan{\frac{(ax+b)}{2}}\right|}+c\)
\(\therefore \int{cosec \ {(ax+b)}dx}=\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c=\frac{1}{a}\ln{\left|\tan{\frac{(ax+b)}{2}}\right|}+c\)
(proved)

\(\int{\sin^{5}{x}\cos^{4}{x}dx}\)
\(=\int{\sin^{4}{x}\cos^{4}{x}\sin{x}dx}\)
\(=\int{(\sin^{2}{x})^2\cos^{4}{x}\sin{x}dx}\)
\(=\int{(1-\cos^{2}{x})^2\cos^{4}{x}\sin{x}dx}\)
\(=\int{(1-t^2)^2t^4(-dt)}\)
\(=-\int{(1-2t^2+t^4)t^4dt}\)
\(=-\int{(t^4-2t^6+t^8)dt}\)
\(=-\int{t^4dt}+2\int{t^6dt}-\int{t^8dt}\)
\(=-\frac{t^{4+1}}{4+1}+2\frac{t^{6+1}}{6+1}-\frac{t^{8+1}}{8+1}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{5}}{5}+2\frac{t^{7}}{7}-\frac{t^{9}}{9}+c\)
\(=-\frac{1}{5}t^{5}+2.\frac{1}{7}t^{7}-\frac{1}{9}t^{9}+c\)
\(=-\frac{1}{5}\cos^{5}{x}+\frac{2}{7}\cos^{7}{x}-\frac{1}{9}\cos^{9}{x}+c\) ➜Note \(\because t=\cos{x}\)

\(\int{\sin^{5}{x}\cos^{3}{x}dx}\)
\(=\int{\sin^{5}{x}\cos^{2}{x}\cos{x}dx}\)
\(=\int{\sin^{5}{x}(1-\sin^{2}{x})\cos{x}dx}\)
\(=\int{t^{5}(1-t^{2})dt}\)
\(=\int{(t^{5}-t^{7})dt}\)
\(=\int{t^{5}dt}-\int{t^{7}dt}\)
\(=\frac{t^{5+1}}{5+1}-\frac{t^{7+1}}{7+1}+c\) ➜Note \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{6}}{6}-\frac{t^{8}}{8}+c\)
\(=\frac{1}{6}t^{6}-\frac{1}{8}t^{8}+c\)
\(=\frac{1}{6}\sin^{6}{x}-\frac{1}{8}\sin^{8}{x}+c\) ➜Note \(\because t=\sin{x}\)

\(\int{\sin^{7}{x}dx}\)
\(=\int{\sin^{6}{x}\sin{x}dx}\)
\(=\int{(\sin^{2}{x})^3\sin{x}dx}\)
\(=\int{(1-\cos^{2}{x})^3\sin{x}dx}\)
\(=\int{(1-t^{2})^3(-dt)}\)
\(=-\int{(1-3t^{2}+3t^4-t^6)dt}\) ➜Note \(\because (a-b)^3=a^3-3a^2b+3ab^2-b^3\)
\(=-\int{dt}+3\int{t^{2}dt}-3\int{t^4dt}+\int{t^6dt}\)
\(=-t+3\frac{t^{3+1}}{3+1}-3\frac{t^{4+1}}{4+1}+\frac{t^{6+1}}{6+1}+c\) ➜Note \(\because \int{dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-t+3\frac{t^{4}}{4}-3\frac{t^{5}}{5}+\frac{t^{7}}{7}+c\)
\(=-t+\frac{3}{4}t^{4}-\frac{3}{5}t^{5}+\frac{1}{7}t^{7}+c\)
\(=-\cos{x}+\frac{3}{4}\cos^{4}{x}-\frac{3}{5}\cos^{5}{x}+\frac{1}{7}cos^{7}{x}+c\) ➜Note \(\because t=\cos{x}\)

\(\int{\sin^{6}{x}dx}\)
\(=\int{(\sin^{2}{x})^3dx}\)
\(=\frac{1}{8}\int{(2\sin^{2}{x})^3dx}\)
\(=\frac{1}{8}\int{(1-\cos{2x})^3dx}\) ➜Note \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{8}\int{(1-3\cos{2x}+3\cos^2{2x}-\cos^3{2x})dx}\) ➜Note \(\because (a-b)^3=a^3-3a^2b+3ab^2-b^3\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{8}\int{\cos^2{2x}dx}-\frac{1}{8}\int{\cos^3{2x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{2\cos^2{2x}dx}-\frac{1}{32}\int{4\cos^3{2x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{(1+\cos{4x})dx}\)\(-\frac{1}{32}\int{(\cos{6x}+3\cos{2x})dx}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}, 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{dx}+\frac{3}{16}\int{\cos{4x}dx}\)\(-\frac{1}{32}\int{\cos{6x}dx}-\frac{3}{32}\int{\cos{2x}dx}\)
\(=\frac{1}{8}x-\frac{3}{8}.\frac{1}{2}\sin{2x}+\frac{3}{16}x+\frac{3}{16}.\frac{1}{4}\sin{4x}\)\(-\frac{1}{32}.\frac{1}{6}\sin{6x}-\frac{3}{32}.\frac{1}{2}\sin{2x}+c\) ➜Note \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{8}x-\frac{3}{16}\sin{2x}+\frac{3}{16}x+\frac{3}{64}\sin{4x}\)\(-\frac{1}{192}\sin{6x}-\frac{3}{64}\sin{2x}+c\)
\(=\frac{1}{8}x+\frac{3}{16}x-\frac{3}{16}\sin{2x}-\frac{3}{64}\sin{2x}\)\(+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{2+3}{16}x+\frac{-12-3}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{5}{16}x+\frac{-15}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{5}{16}x-\frac{15}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{1}{192}(60x-45\sin{2x}+9\sin{4x}-\sin{6x})+c\)

\(\int{\cos^{5}{x}dx}\)
\(=\int{\cos^{4}{x}\cos{x}dx}\)
\(=\int{(\cos^{2}{x})^2\cos{x}dx}\)
\(=\int{(1-\sin^{2}{x})^2\cos{x}dx}\)
\(=\int{(1-t^{2})^2dt}\)
\(=\int{(1-2t^{2}+t^4)dt}\) ➜Note \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\int{dt}-2\int{t^{2}dt}+\int{t^4dt}\)
\(=t-2\frac{t^{2+1}}{2+1}+\frac{t^{4+1}}{4+1}+c\) ➜Note \(\because \int{dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=t-2\frac{t^{3}}{3}+\frac{t^{5}}{5}+c\)
\(=t-\frac{2}{3}t^{3}+\frac{1}{5}t^{5}+c\)
\(=\sin{x}-\frac{2}{3}\sin^{3}{x}+\frac{1}{5}\sin^{5}{x}+c\) ➜Note \(\because t=\sin{x}\)

\(\int{\cos^{4}{x}dx}\)
\(=\int{(\cos^{2}{x})^2dx}\)
\(=\frac{1}{4}\int{(2\cos^{2}{x})^2dx}\)
\(=\frac{1}{4}\int{(1+\cos{2x})^2dx}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜Note \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜Note \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}x+\frac{1}{2}.\frac{1}{2}\sin{2x}+\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}\sin{4x}+c\) ➜Note \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{2+1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)