Class 12: Indefinite Integrals

$\displaystyle \textbf{Question: }~\text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 1: }~\int \frac{\log x}{x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\log x}{x}\,dx$

$\displaystyle \text{Let } \log x = t$

$\displaystyle \Rightarrow \frac{dt}{dx} = \frac{1}{x}$

$\displaystyle \Rightarrow dt = \frac{1}{x}\,dx$

$\displaystyle \text{Now, } \int \frac{\log x}{x}\,dx = \int t\,dt$

$\displaystyle = \frac{t^{2}}{2} + C$

$\displaystyle = \frac{(\log x)^{2}}{2} + C$

$\displaystyle \textbf{Question 2: }~\int \frac{\log\!\left(1+\frac{1}{x}\right)}{x(1+x)}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\log\!\left(1+\frac{1}{x}\right)}{x(1+x)}\,dx$

$\displaystyle \text{Let } \log\!\left(1+\frac{1}{x}\right)=t$

$\displaystyle \Rightarrow \frac{1}{1+\frac{1}{x}}\cdot\left(-\frac{1}{x^{2}}\right)=\frac{dt}{dx}$

$\displaystyle \Rightarrow \frac{x}{x+1}\cdot\left(-\frac{1}{x^{2}}\right)=\frac{dt}{dx}$

$\displaystyle \Rightarrow -\frac{1}{x(x+1)}=\frac{dt}{dx}$

$\displaystyle \Rightarrow -\frac{dx}{x(x+1)}=dt$

$\displaystyle \Rightarrow \frac{dx}{x(x+1)}=-dt$

$\displaystyle \text{Now, } \int \frac{\log\!\left(1+\frac{1}{x}\right)}{x(1+x)}\,dx=\int t(-dt)$

$\displaystyle =-\frac{t^{2}}{2}+C$

$\displaystyle =-\frac{1}{2}\left\{\log\!\left(1+\frac{1}{x}\right)\right\}^{2}+C$

$\displaystyle \textbf{Question 3: }~\int \frac{(1+\sqrt{x})^2}{\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{(1+\sqrt{x})^{2}}{\sqrt{x}}\,dx$

$\displaystyle \text{Let } 1+\sqrt{x}=t$

$\displaystyle \Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$

$\displaystyle \text{Now, } \int \frac{(1+\sqrt{x})^{2}}{\sqrt{x}}\,dx = \int t^{2}\,(2\,dt)$

$\displaystyle = 2\int t^{2}\,dt$

$\displaystyle = \frac{2}{3}t^{3}+C$

$\displaystyle = \frac{2}{3}(1+\sqrt{x})^{3}+C$

$\displaystyle \textbf{Question 4: }~\int \sqrt{1+e^x}\,e^x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \sqrt{1+e^{x}}\cdot e^{x}\,dx$

$\displaystyle \text{Let } 1+e^{x}=t$

$\displaystyle \Rightarrow e^{x}=\frac{dt}{dx}$

$\displaystyle \Rightarrow e^{x}\,dx=dt$

$\displaystyle \text{Now, } \int \sqrt{1+e^{x}}\cdot e^{x}\,dx=\int \sqrt{t}\,dt$

$\displaystyle =\int t^{\frac{1}{2}}\,dt$

$\displaystyle =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C$

$\displaystyle =\frac{2}{3}t^{\frac{3}{2}}+C$

$\displaystyle =\frac{2}{3}(1+e^{x})^{\frac{3}{2}}+C$

$\displaystyle \textbf{Question 5: }~\int \sqrt[3]{\cos^2 x}\,\sin x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int (\cos^{2}x)^{\frac{1}{3}}\sin x\,dx$

$\displaystyle \text{Let } \cos x=t$

$\displaystyle \Rightarrow -\sin x=\frac{dt}{dx}$

$\displaystyle \Rightarrow \sin x\,dx=-dt$

$\displaystyle \text{Now, } \int (\cos^{2}x)^{\frac{1}{3}}\sin x\,dx = -\int t^{\frac{2}{3}}\,dt$

$\displaystyle = -\frac{t^{\frac{5}{3}}}{\frac{5}{3}}+C$

$\displaystyle = -\frac{3}{5}t^{\frac{5}{3}}+C$

$\displaystyle = -\frac{3}{5}\cos^{\frac{5}{3}}x+C$

$\displaystyle \textbf{Question 6: }~\int \frac{e^x}{(1+e^x)^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{e^{x}}{(1+e^{x})^{2}}\,dx$

$\displaystyle \text{Let } 1+e^{x}=t$

$\displaystyle \Rightarrow e^{x}=\frac{dt}{dx}$

$\displaystyle \Rightarrow e^{x}\,dx=dt$

$\displaystyle \text{Now, } \int \frac{e^{x}}{(1+e^{x})^{2}}\,dx=\int \frac{dt}{t^{2}}$

$\displaystyle =\int t^{-2}\,dt$

$\displaystyle =\frac{t^{-1}}{-1}+C$

$\displaystyle =-\frac{1}{t}+C$

$\displaystyle =-\frac{1}{1+e^{x}}+C$

$\displaystyle \textbf{Question 7: }~\int \cot^3 x~\mathrm{cosec}^2 x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \cot^{3}x\,\mathrm{cosec}^{2}x\,dx$

$\displaystyle \text{Let } \cot x=t$

$\displaystyle \Rightarrow -\mathrm{cosec}^{2}x=\frac{dt}{dx}$

$\displaystyle \Rightarrow \mathrm{cosec}^{2}x\,dx=-dt$

$\displaystyle \text{Now, } \int \cot^{3}x\,\mathrm{cosec}^{2}x\,dx=\int t^{3}(-dt)$

$\displaystyle =-\int t^{3}\,dt$

$\displaystyle =-\frac{t^{4}}{4}+C$

$\displaystyle =-\frac{\cot^{4}x}{4}+C$

$\displaystyle \textbf{Question 8: }~\int \frac{\left(e^{\sin^{-1}x}\right)^2}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\left(e^{\sin^{-1}x}\right)^{2}}{\sqrt{1-x^{2}}}\,dx$

$\displaystyle \text{Let } e^{\sin^{-1}x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=e^{\sin^{-1}x}\cdot\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle \Rightarrow dt=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\,dx$

$\displaystyle \text{Now, } \int \frac{\left(e^{\sin^{-1}x}\right)^{2}}{\sqrt{1-x^{2}}}\,dx=\int e^{\sin^{-1}x}\cdot\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\,dx$

$\displaystyle =\int t\,dt$

$\displaystyle =\frac{t^{2}}{2}+C$

$\displaystyle =\frac{\left(e^{\sin^{-1}x}\right)^{2}}{2}+C$

$\displaystyle \textbf{Question 9: }~\int \frac{1+\sin x}{\sqrt{x-\cos x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{1+\sin x}{\sqrt{x-\cos x}}\,dx$

$\displaystyle \text{Let } x-\cos x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=1+\sin x$

$\displaystyle \Rightarrow (1+\sin x)\,dx=dt$

$\displaystyle \text{Now, } \int \frac{1+\sin x}{\sqrt{x-\cos x}}\,dx=\int \frac{dt}{\sqrt{t}}$

$\displaystyle =\int t^{-\frac{1}{2}}\,dt$

$\displaystyle =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C$

$\displaystyle =2\sqrt{t}+C$

$\displaystyle =2\sqrt{x-\cos x}+C$

$\displaystyle \textbf{Question 10: }~\int \frac{1}{\sqrt{1-x^2}\,(\sin^{-1}x)^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{dx}{\sqrt{1-x^{2}}\,(\sin^{-1}x)^{2}}$

$\displaystyle \text{Let } \sin^{-1}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{1-x^{2}}}=dt$

$\displaystyle \text{Now, } \int \frac{dx}{\sqrt{1-x^{2}}\,(\sin^{-1}x)^{2}}=\int \frac{dt}{t^{2}}$

$\displaystyle =\int t^{-2}\,dt$

$\displaystyle =\frac{t^{-1}}{-1}+C$

$\displaystyle =-\frac{1}{t}+C$

$\displaystyle =-\frac{1}{\sin^{-1}x}+C$

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 11: }~\int \frac{\cot x}{\sqrt{\sin x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\cot x}{\sqrt{\sin x}}\,dx$

$\displaystyle = \int \frac{\cos x}{\sin x\,\sqrt{\sin x}}\,dx$

$\displaystyle = \int \frac{\cos x}{(\sin x)^{\frac{3}{2}}}\,dx$

$\displaystyle \text{Let } \sin x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\cos x$

$\displaystyle \Rightarrow \cos x\,dx=dt$

$\displaystyle \text{Now, } \int \frac{\cos x}{(\sin x)^{\frac{3}{2}}}\,dx=\int \frac{dt}{t^{\frac{3}{2}}}$

$\displaystyle =\int t^{-\frac{3}{2}}\,dt$

$\displaystyle =\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}+C$

$\displaystyle =-\frac{2}{\sqrt{t}}+C$

$\displaystyle =-\frac{2}{\sqrt{\sin x}}+C$

$\displaystyle \textbf{Question 12: }~\int \frac{\tan x}{\sqrt{\cos x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\tan x}{\sqrt{\cos x}}\,dx$

$\displaystyle = \int \frac{\sin x}{\cos x\,\sqrt{\cos x}}\,dx$

$\displaystyle = \int \frac{\sin x}{(\cos x)^{\frac{3}{2}}}\,dx$

$\displaystyle \text{Let } \cos x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-\sin x$

$\displaystyle \Rightarrow \sin x\,dx=-dt$

$\displaystyle \text{Now, } \int \frac{\sin x}{(\cos x)^{\frac{3}{2}}}\,dx=-\int \frac{dt}{t^{\frac{3}{2}}}$

$\displaystyle =-\int t^{-\frac{3}{2}}\,dt$

$\displaystyle =-\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}+C$

$\displaystyle =\frac{2}{\sqrt{t}}+C$

$\displaystyle =\frac{2}{\sqrt{\cos x}}+C$

$\displaystyle \textbf{Question 13: }~\int \frac{\cos^3 x}{\sqrt{\sin x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\cos^{3}x}{\sqrt{\sin x}}\,dx$

$\displaystyle = \int \frac{\cos^{2}x\cdot \cos x}{\sqrt{\sin x}}\,dx$

$\displaystyle = \int \frac{(1-\sin^{2}x)\cos x}{\sqrt{\sin x}}\,dx$

$\displaystyle \text{Let } \sin x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\cos x$

$\displaystyle \Rightarrow \cos x\,dx=dt$

$\displaystyle \text{Now, } \int \frac{(1-\sin^{2}x)\cos x}{\sqrt{\sin x}}\,dx=\int \frac{1-t^{2}}{\sqrt{t}}\,dt$

$\displaystyle =\int \left(t^{-\frac{1}{2}}-t^{\frac{3}{2}}\right)dt$

$\displaystyle =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}-\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+C$

$\displaystyle =2\sqrt{t}-\frac{2}{5}t^{\frac{5}{2}}+C$

$\displaystyle =2\sqrt{\sin x}-\frac{2}{5}\sin^{\frac{5}{2}}x+C$

$\displaystyle \textbf{Question 14: }~\int \frac{\sin^3 x}{\sqrt{\cos x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin^{3}x}{\sqrt{\cos x}}\,dx$

$\displaystyle = \int \frac{\sin^{2}x\cdot \sin x}{\sqrt{\cos x}}\,dx$

$\displaystyle = \int \frac{(1-\cos^{2}x)\sin x}{\sqrt{\cos x}}\,dx$

$\displaystyle \text{Let } \cos x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-\sin x$

$\displaystyle \Rightarrow \sin x\,dx=-dt$

$\displaystyle \text{Now, } \int \frac{(1-\cos^{2}x)\sin x}{\sqrt{\cos x}}\,dx=-\int \frac{1-t^{2}}{\sqrt{t}}\,dt$

$\displaystyle =\int \left(t^{\frac{3}{2}}-t^{-\frac{1}{2}}\right)dt$

$\displaystyle =\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C$

$\displaystyle =\frac{2}{5}t^{\frac{5}{2}}-2\sqrt{t}+C$

$\displaystyle =\frac{2}{5}\cos^{\frac{5}{2}}x-2\sqrt{\cos x}+C$

$\displaystyle \textbf{Question 15: }~\int \frac{1}{\sqrt{\tan^{-1}x}\,(1+x^2)}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{dx}{\sqrt{\tan^{-1}x}\,(1+x^{2})}$

$\displaystyle \text{Let } \tan^{-1}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{1+x^{2}}$

$\displaystyle \Rightarrow \frac{dx}{1+x^{2}}=dt$

$\displaystyle \text{Now, } \int \frac{dx}{\sqrt{\tan^{-1}x}\,(1+x^{2})}=\int \frac{dt}{\sqrt{t}}$

$\displaystyle =\int t^{-\frac{1}{2}}\,dt$

$\displaystyle =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C$

$\displaystyle =2\sqrt{t}+C$

$\displaystyle =2\sqrt{\tan^{-1}x}+C$

$\displaystyle \textbf{Question 16: }~\int \frac{\sqrt{\tan x}}{\sin x\cos x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sqrt{\tan x}}{\sin x \cos x}\,dx$

$\displaystyle = \int \frac{\sqrt{\tan x}}{\dfrac{\sin x}{\cos x}\,\cos^{2}x}\,dx$

$\displaystyle = \int \frac{\sqrt{\tan x}}{\tan x}\,\sec^{2}x\,dx$

$\displaystyle = \int \frac{1}{\sqrt{\tan x}}\,\sec^{2}x\,dx$

$\displaystyle = \int (\tan x)^{-\frac{1}{2}}\,\sec^{2}x\,dx$

$\displaystyle \text{Let } \tan x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\sec^{2}x$

$\displaystyle \Rightarrow \sec^{2}x\,dx=dt$

$\displaystyle \text{Now, } \int (\tan x)^{-\frac{1}{2}}\,\sec^{2}x\,dx=\int t^{-\frac{1}{2}}\,dt$

$\displaystyle =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C$

$\displaystyle =2\sqrt{t}+C$

$\displaystyle =2\sqrt{\tan x}+C$

$\displaystyle \textbf{Question 17: }~\int x(\log x)^2\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{1}{x}(\log x)^{2}\,dx$

$\displaystyle \text{Let } \log x=t$

$\displaystyle \Rightarrow \frac{1}{x}\,dx=dt$

$\displaystyle \text{Now, } \int \frac{1}{x}(\log x)^{2}\,dx=\int t^{2}\,dt$

$\displaystyle =\frac{t^{3}}{3}+C$

$\displaystyle =\frac{(\log x)^{3}}{3}+C$

$\displaystyle \textbf{Question 18: }~\int \sin^5 x\cos x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \sin^{5}x \cos x\,dx$

$\displaystyle \text{Let } \sin x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\cos x$

$\displaystyle \Rightarrow \cos x\,dx=dt$

$\displaystyle \text{Now, } \int \sin^{5}x \cos x\,dx=\int t^{5}\,dt$

$\displaystyle =\frac{t^{6}}{6}+C$

$\displaystyle =\frac{1}{6}\sin^{6}x+C$

$\displaystyle \textbf{Question 19: }~\int \tan^{3/2}x\,\sec^2 x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \tan^{\frac{3}{2}}x \cdot \sec^{2}x\,dx$

$\displaystyle \text{Let } \tan x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\sec^{2}x$

$\displaystyle \Rightarrow \sec^{2}x\,dx=dt$

$\displaystyle \text{Now, } \int \tan^{\frac{3}{2}}x \cdot \sec^{2}x\,dx=\int t^{\frac{3}{2}}\,dt$

$\displaystyle =\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+C$

$\displaystyle =\frac{2}{5}t^{\frac{5}{2}}+C$

$\displaystyle =\frac{2}{5}\tan^{\frac{5}{2}}x+C$

$\displaystyle \textbf{Question 20: }~\int \frac{x^3}{(x^2+3)^3}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{x^{3}}{(x^{2}+1)^{3}}\,dx$

$\displaystyle = \int \frac{x^{2}\cdot x}{(x^{2}+1)^{3}}\,dx$

$\displaystyle \text{Let } x^{2}+1=t$

$\displaystyle \Rightarrow 2x\,dx=dt$

$\displaystyle \Rightarrow x\,dx=\frac{dt}{2}$

$\displaystyle \text{Now, } \int \frac{x^{2}\cdot x}{(x^{2}+1)^{3}}\,dx=\frac{1}{2}\int \frac{t-1}{t^{3}}\,dt$

$\displaystyle =\frac{1}{2}\int \left(t^{-2}-t^{-3}\right)dt$

$\displaystyle =\frac{1}{2}\left(\frac{t^{-1}}{-1}-\frac{t^{-2}}{-2}\right)+C$

$\displaystyle =\frac{1}{2}\left(-\frac{1}{t}+\frac{1}{2t^{2}}\right)+C$

$\displaystyle =-\frac{1}{2(x^{2}+1)}+\frac{1}{4(x^{2}+1)^{2}}+C$

$\displaystyle =-\frac{2x^{2}+1}{4(x^{2}+1)^{2}}+C$

$\displaystyle \textbf{Question 21: }~\int (4x+2)\sqrt{x^2+x+1}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int (4x+2)\sqrt{x^{2}+x+1}\,dx$

$\displaystyle =2\int (2x+1)\sqrt{x^{2}+x+1}\,dx$

$\displaystyle \text{Let } x^{2}+x+1=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=2x+1$

$\displaystyle \Rightarrow (2x+1)\,dx=dt$

$\displaystyle \text{Now, } 2\int (2x+1)\sqrt{x^{2}+x+1}\,dx=2\int \sqrt{t}\,dt$

$\displaystyle =2\int t^{\frac{1}{2}}\,dt$

$\displaystyle =2\cdot\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C$

$\displaystyle =\frac{4}{3}t^{\frac{3}{2}}+C$

$\displaystyle =\frac{4}{3}(x^{2}+x+1)^{\frac{3}{2}}+C$

$\displaystyle \textbf{Question 22: }~\int \frac{4x+3}{\sqrt{2x^2+3x+1}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{4x+3}{\sqrt{2x^{2}+3x+1}}\,dx$

$\displaystyle \text{Let } 2x^{2}+3x+1=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=4x+3$

$\displaystyle \Rightarrow (4x+3)\,dx=dt$

$\displaystyle \text{Now, } \int \frac{4x+3}{\sqrt{2x^{2}+3x+1}}\,dx=\int \frac{dt}{\sqrt{t}}$

$\displaystyle =\int t^{-\frac{1}{2}}\,dt$

$\displaystyle =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C$

$\displaystyle =2\sqrt{t}+C$

$\displaystyle =2\sqrt{2x^{2}+3x+1}+C$

$\displaystyle \textbf{Question 23: }~\int \frac{1}{1+\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{dx}{1+\sqrt{x}}$

$\displaystyle =\int \frac{\sqrt{x}}{\sqrt{x}(1+\sqrt{x})}\,dx$

$\displaystyle \text{Let } 1+\sqrt{x}=t$

$\displaystyle \Rightarrow \sqrt{x}=t-1$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$

$\displaystyle \text{Now, } \int \frac{\sqrt{x}}{\sqrt{x}(1+\sqrt{x})}\,dx=\int \frac{\sqrt{x}}{t}\,dx$

$\displaystyle =\int \frac{t-1}{t}\,dx$

$\displaystyle =\int \frac{t-1}{t}\,(2\,dt)$

$\displaystyle =2\int \left(1-\frac{1}{t}\right)dt$

$\displaystyle =2\left(t-\log|t|\right)+C$

$\displaystyle =2\left(1+\sqrt{x}-\log(1+\sqrt{x})\right)+C$

$\displaystyle =2\sqrt{x}-2\log(1+\sqrt{x})+C$

$\displaystyle \textbf{Question 24: }~\int e^{\cos^2 x}\sin 2x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } I=\int e^{\cos^{2}x}\,\sin 2x\,dx$

$\displaystyle \text{Using } \sin 2x=2\sin x\cos x$

$\displaystyle I=\int e^{\cos^{2}x}\cdot 2\sin x\cos x\,dx$

$\displaystyle \text{Let } \cos^{2}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-2\sin x\cos x$

$\displaystyle \Rightarrow -2\sin x\cos x\,dx=dt$

$\displaystyle \text{Now, } I=\int e^{t}\,(-dt)$

$\displaystyle =-\int e^{t}\,dt$

$\displaystyle =-e^{t}+C$

$\displaystyle =-e^{\cos^{2}x}+C$

$\displaystyle \textbf{Question 25: }~\int \frac{1+\cos x}{(x+\sin x)^3}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{1+\cos x}{(x+\sin x)^{3}}\,dx$

$\displaystyle \text{Let } x+\sin x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=1+\cos x$

$\displaystyle \Rightarrow (1+\cos x)\,dx=dt$

$\displaystyle \text{Now, } \int \frac{1+\cos x}{(x+\sin x)^{3}}\,dx=\int \frac{dt}{t^{3}}$

$\displaystyle =\int t^{-3}\,dt$

$\displaystyle =\frac{t^{-2}}{-2}+C$

$\displaystyle =-\frac{1}{2t^{2}}+C$

$\displaystyle =-\frac{1}{2(x+\sin x)^{2}}+C$

$\displaystyle \textbf{Question 26: }~\int \frac{\cos x-\sin x}{1+\sin 2x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\cos x-\sin x}{1+\sin 2x}\,dx$

$\displaystyle =\int \frac{\cos x-\sin x}{\cos^{2}x+\sin^{2}x+2\sin x\cos x}\,dx$

$\displaystyle =\int \frac{\cos x-\sin x}{(\cos x+\sin x)^{2}}\,dx$

$\displaystyle \text{Let } \cos x+\sin x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-\sin x+\cos x$

$\displaystyle \Rightarrow (\cos x-\sin x)\,dx=dt$

$\displaystyle \text{Now, } \int \frac{\cos x-\sin x}{(\cos x+\sin x)^{2}}\,dx=\int \frac{dt}{t^{2}}$

$\displaystyle =\int t^{-2}\,dt$

$\displaystyle =\frac{t^{-1}}{-1}+C$

$\displaystyle =-\frac{1}{t}+C$

$\displaystyle =-\frac{1}{\sin x+\cos x}+C$

$\displaystyle \textbf{Question 27: }~\int \frac{\sin 2x}{(a+b\cos 2x)^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin(2x)}{(a+b\cos 2x)^{2}}\,dx$

$\displaystyle \text{Let } a+b\cos 2x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-2b\sin(2x)$

$\displaystyle \Rightarrow \sin(2x)\,dx=-\frac{dt}{2b}$

$\displaystyle \text{Now, } \int \frac{\sin(2x)}{(a+b\cos 2x)^{2}}\,dx=-\frac{1}{2b}\int \frac{dt}{t^{2}}$

$\displaystyle =-\frac{1}{2b}\int t^{-2}\,dt$

$\displaystyle =-\frac{1}{2b}\left(\frac{t^{-1}}{-1}\right)+C$

$\displaystyle =\frac{1}{2b}\cdot\frac{1}{t}+C$

$\displaystyle =\frac{1}{2b(a+b\cos 2x)}+C$

$\displaystyle \textbf{Question 28: }~\int \frac{\log x^2}{x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\log x^{2}}{x}\,dx$

$\displaystyle =\int \frac{2\log x}{x}\,dx$

$\displaystyle =2\int \frac{\log x}{x}\,dx$

$\displaystyle \text{Let } \log x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{x}$

$\displaystyle \Rightarrow \frac{1}{x}\,dx=dt$

$\displaystyle \text{Now, } 2\int \frac{\log x}{x}\,dx=2\int t\,dt$

$\displaystyle =2\left(\frac{t^{2}}{2}\right)+C$

$\displaystyle =t^{2}+C$

$\displaystyle =(\log x)^{2}+C$

$\displaystyle \textbf{Question 29: }~\int \frac{\sin x}{(1+\cos x)^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin x}{(1+\cos x)^{2}}\,dx$

$\displaystyle \text{Let } 1+\cos x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-\sin x$

$\displaystyle \Rightarrow \sin x\,dx=-dt$

$\displaystyle \text{Now, } \int \frac{\sin x}{(1+\cos x)^{2}}\,dx=-\int \frac{dt}{t^{2}}$

$\displaystyle =-\int t^{-2}\,dt$

$\displaystyle =-\left(\frac{t^{-1}}{-1}\right)+C$

$\displaystyle =\frac{1}{t}+C$

$\displaystyle =\frac{1}{1+\cos x}+C$

$\displaystyle \textbf{Question 30: }~\int \cot x\log\sin x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \cot x\cdot \log(\sin x)\,dx$

$\displaystyle \text{Let } \log(\sin x)=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\sin x}\cdot \cos x$

$\displaystyle \Rightarrow \cot x\,dx=dt$

$\displaystyle \text{Now, } \int \cot x\cdot \log(\sin x)\,dx=\int t\,dt$

$\displaystyle =\frac{t^{2}}{2}+C$

$\displaystyle =\frac{(\log(\sin x))^{2}}{2}+C$

$\displaystyle \textbf{Question 31: }~\int \sec x\log(\sec x+\tan x)\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \sec x\cdot \log(\sec x+\tan x)\,dx$

$\displaystyle \text{Let } \log(\sec x+\tan x)=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{\sec x\tan x+\sec^{2}x}{\sec x+\tan x}$

$\displaystyle \Rightarrow \frac{dt}{dx}=\sec x$

$\displaystyle \Rightarrow \sec x\,dx=dt$

$\displaystyle \text{Now, } \int \sec x\cdot \log(\sec x+\tan x)\,dx=\int t\,dt$

$\displaystyle =\frac{t^{2}}{2}+C$

$\displaystyle =\frac{\left(\log(\sec x+\tan x)\right)^{2}}{2}+C$

$\displaystyle \textbf{Question 32: }~\int \mathrm{cosec}\,x\log(\mathrm{cosec}\,x-\cot x)\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \mathrm{cosec}\,x\cdot \log(\mathrm{cosec}\,x-\cot x)\,dx$

$\displaystyle \text{Let } \log(\mathrm{cosec}\,x-\cot x)=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{-\mathrm{cosec}\,x\cot x+\mathrm{cosec}^{2}x}{\mathrm{cosec}\,x-\cot x}$

$\displaystyle \Rightarrow \frac{dt}{dx}=\mathrm{cosec}\,x$

$\displaystyle \Rightarrow \mathrm{cosec}\,x\,dx=dt$

$\displaystyle \text{Now, } \int \mathrm{cosec}\,x\cdot \log(\mathrm{cosec}\,x-\cot x)\,dx=\int t\,dt$

$\displaystyle =\frac{t^{2}}{2}+C$

$\displaystyle =\frac{\left(\log(\mathrm{cosec}\,x-\cot x)\right)^{2}}{2}+C$

$\displaystyle \textbf{Question 33: }~\int x^3\cos x^4\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int x^{3}\cos(x^{4})\,dx$

$\displaystyle \text{Let } x^{4}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=4x^{3}$

$\displaystyle \Rightarrow x^{3}\,dx=\frac{dt}{4}$

$\displaystyle \text{Now, } \int x^{3}\cos(x^{4})\,dx=\frac{1}{4}\int \cos t\,dt$

$\displaystyle =\frac{1}{4}\sin t+C$

$\displaystyle =\frac{1}{4}\sin(x^{4})+C$

$\displaystyle \textbf{Question 34: }~\int x^3\sin x^4\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int x^{3}\sin(x^{4})\,dx$

$\displaystyle \text{Let } x^{4}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=4x^{3}$

$\displaystyle \Rightarrow x^{3}\,dx=\frac{dt}{4}$

$\displaystyle \text{Now, } \int x^{3}\sin(x^{4})\,dx=\frac{1}{4}\int \sin t\,dt$

$\displaystyle =\frac{1}{4}(-\cos t)+C$

$\displaystyle =-\frac{1}{4}\cos(x^{4})+C$

$\displaystyle \textbf{Question 35: }~\int \frac{x\sin^{-1}(x^2)}{\sqrt{1-x^4}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{x\sin^{-1}x^{2}}{\sqrt{1-x^{4}}}\,dx$

$\displaystyle \text{Let } \sin^{-1}x^{2}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{2x}{\sqrt{1-x^{4}}}$

$\displaystyle \Rightarrow \frac{x\,dx}{\sqrt{1-x^{4}}}=\frac{dt}{2}$

$\displaystyle \text{Now, } \int \frac{x\sin^{-1}x^{2}}{\sqrt{1-x^{4}}}\,dx=\frac{1}{2}\int t\,dt$

$\displaystyle =\frac{t^{2}}{4}+C$

$\displaystyle =\frac{(\sin^{-1}x^{2})^{2}}{4}+C$

$\displaystyle \textbf{Question 36: }~\int x^3\sin(x^4+1)\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int x^{3}\sin(x^{4}+1)\,dx$

$\displaystyle \text{Let } x^{4}+1=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=4x^{3}$

$\displaystyle \Rightarrow x^{3}\,dx=\frac{dt}{4}$

$\displaystyle \text{Now, } \int x^{3}\sin(x^{4}+1)\,dx=\frac{1}{4}\int \sin t\,dt$

$\displaystyle =-\frac{1}{4}\cos t+C$

$\displaystyle =-\frac{1}{4}\cos(x^{4}+1)+C$

$\displaystyle \textbf{Question 37: }~\int \frac{(x+1)e^x}{\cos^2(e^x)}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{(x+1)e^{x}}{\cos^{2}(x e^{x})}\,dx$

$\displaystyle \text{Let } x e^{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=e^{x}+x e^{x}$

$\displaystyle \Rightarrow (x+1)e^{x}\,dx=dt$

$\displaystyle \text{Now, } \int \frac{(x+1)e^{x}}{\cos^{2}(x e^{x})}\,dx=\int \frac{dt}{\cos^{2}t}$

$\displaystyle =\int \sec^{2}t\,dt$

$\displaystyle =\tan t+C$

$\displaystyle =\tan(x e^{x})+C$

$\displaystyle \textbf{Question 38: }~\int x^2 e^{x^3}\cos(e^{x^3})\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int x^{2}e^{x^{3}}\cos\!\left(e^{x^{3}}\right)\,dx$

$\displaystyle \text{Let } e^{x^{3}}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=e^{x^{3}}\cdot 3x^{2}$

$\displaystyle \Rightarrow e^{x^{3}}\,x^{2}\,dx=\frac{dt}{3}$

$\displaystyle \text{Now, } \int x^{2}e^{x^{3}}\cos\!\left(e^{x^{3}}\right)\,dx=\frac{1}{3}\int \cos t\,dt$

$\displaystyle =\frac{1}{3}\sin t+C$

$\displaystyle =\frac{1}{3}\sin\!\left(e^{x^{3}}\right)+C$

$\displaystyle \textbf{Question 39: }~\int 2x\,\sec^3(x^2+3)\tan(x^2+3)\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int 2x\,\sec^{3}(x^{2}+3)\,\tan(x^{2}+3)\,dx$

$\displaystyle =\int \sec^{2}(x^{2}+3)\cdot \sec(x^{2}+3)\tan(x^{2}+3)\cdot 2x\,dx$

$\displaystyle \text{Let } \sec(x^{2}+3)=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\sec(x^{2}+3)\tan(x^{2}+3)\cdot 2x$

$\displaystyle \Rightarrow \sec(x^{2}+3)\tan(x^{2}+3)\cdot 2x\,dx=dt$

$\displaystyle \text{Now, } \int \sec^{2}(x^{2}+3)\cdot \sec(x^{2}+3)\tan(x^{2}+3)\cdot 2x\,dx=\int t^{2}\,dt$

$\displaystyle =\frac{t^{3}}{3}+C$

$\displaystyle =\frac{\sec^{3}(x^{2}+3)}{3}+C$

$\displaystyle \textbf{Question 40: }~\int \left(\frac{x+1}{x}\right)(x+\log x)^2\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \left(\frac{x+1}{x}\right)(x+\log x)^{2}\,dx$

$\displaystyle \text{Let } x+\log x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=1+\frac{1}{x}$

$\displaystyle \Rightarrow \frac{x+1}{x}\,dx=dt$

$\displaystyle \text{Now, } \int \left(\frac{x+1}{x}\right)(x+\log x)^{2}\,dx=\int t^{2}\,dt$

$\displaystyle =\frac{t^{3}}{3}+C$

$\displaystyle =\frac{(x+\log x)^{3}}{3}+C$

$\displaystyle \textbf{Question 41: }~\int \tan x\sec^2 x\sqrt{1-\tan^2 x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \tan x\cdot \sec^{2}x\,\sqrt{1-\tan^{2}x}\,dx$

$\displaystyle \text{Let } \tan x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\sec^{2}x$

$\displaystyle \Rightarrow \sec^{2}x\,dx=dt$

$\displaystyle \text{Now, } \int \tan x\cdot \sec^{2}x\,\sqrt{1-\tan^{2}x}\,dx=\int t\sqrt{1-t^{2}}\,dt$

$\displaystyle \text{Let } 1-t^{2}=p$

$\displaystyle \Rightarrow \frac{dp}{dt}=-2t$

$\displaystyle \Rightarrow t\,dt=-\frac{dp}{2}$

$\displaystyle \text{Now, } \int t\sqrt{1-t^{2}}\,dt=-\frac{1}{2}\int \sqrt{p}\,dp$

$\displaystyle =-\frac{1}{2}\int p^{\frac{1}{2}}\,dp$

$\displaystyle =-\frac{1}{2}\cdot\frac{p^{\frac{3}{2}}}{\frac{3}{2}}+C$

$\displaystyle =-\frac{1}{3}p^{\frac{3}{2}}+C$

$\displaystyle =-\frac{1}{3}(1-t^{2})^{\frac{3}{2}}+C$

$\displaystyle =-\frac{1}{3}(1-\tan^{2}x)^{\frac{3}{2}}+C$

$\displaystyle \textbf{Question 42: }~\int \log x\frac{\sin\{1+(\log x)^2\}}{x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\log x\,\sin\!\left(1+(\log x)^{2}\right)}{x}\,dx$

$\displaystyle \text{Let } 1+(\log x)^{2}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=2\log x\cdot \frac{1}{x}$

$\displaystyle \Rightarrow \frac{\log x}{x}\,dx=\frac{dt}{2}$

$\displaystyle \text{Now, } \int \frac{\log x\,\sin\!\left(1+(\log x)^{2}\right)}{x}\,dx=\frac{1}{2}\int \sin t\,dt$

$\displaystyle =-\frac{1}{2}\cos t+C$

$\displaystyle =-\frac{1}{2}\cos\!\left(1+(\log x)^{2}\right)+C$

$\displaystyle \textbf{Question 43: }~\int \frac{1}{x^2}\cos^2\!\left(\frac{1}{x}\right)\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{1}{x^{2}}\cos^{2}\!\left(\frac{1}{x}\right)\,dx$

$\displaystyle \text{Let } \frac{1}{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-\frac{1}{x^{2}}$

$\displaystyle \Rightarrow \frac{1}{x^{2}}\,dx=-dt$

$\displaystyle \text{Now, } \int \frac{1}{x^{2}}\cos^{2}\!\left(\frac{1}{x}\right)\,dx=-\int \cos^{2}t\,dt$

$\displaystyle =-\int \frac{1+\cos 2t}{2}\,dt$

$\displaystyle =-\frac{1}{2}\int (1+\cos 2t)\,dt$

$\displaystyle =-\frac{1}{2}\left(t+\frac{\sin 2t}{2}\right)+C$

$\displaystyle =-\frac{1}{2}\left(\frac{1}{x}+\frac{\sin\!\left(\frac{2}{x}\right)}{2}\right)+C$

$\displaystyle =-\frac{1}{2x}-\frac{1}{4}\sin\!\left(\frac{2}{x}\right)+C$

$\displaystyle \textbf{Question 44: }~\int \sec^4 x\tan x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \sec^{4}x\cdot \tan x\,dx$

$\displaystyle =\int \sec^{2}x\cdot \sec^{2}x\cdot \tan x\,dx$

$\displaystyle =\int (1+\tan^{2}x)\sec^{2}x\cdot \tan x\,dx$

$\displaystyle =\int (\tan x+\tan^{3}x)\sec^{2}x\,dx$

$\displaystyle \text{Let } \tan x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\sec^{2}x$

$\displaystyle \Rightarrow \sec^{2}x\,dx=dt$

$\displaystyle \text{Now, } \int (\tan x+\tan^{3}x)\sec^{2}x\,dx=\int (t+t^{3})\,dt$

$\displaystyle =\frac{t^{2}}{2}+\frac{t^{4}}{4}+C$

$\displaystyle =\frac{\tan^{2}x}{2}+\frac{\tan^{4}x}{4}+C$

$\displaystyle \textbf{Question 45: }~\int \frac{e^{\sqrt{x}}\cos(e^{\sqrt{x}})}{\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{e^{\sqrt{x}}\cos\!\left(e^{\sqrt{x}}\right)}{\sqrt{x}}\,dx$

$\displaystyle \text{Let } e^{\sqrt{x}}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}$

$\displaystyle \Rightarrow \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx=2\,dt$

$\displaystyle \text{Now, } \int \frac{e^{\sqrt{x}}\cos\!\left(e^{\sqrt{x}}\right)}{\sqrt{x}}\,dx=2\int \cos t\,dt$

$\displaystyle =2\sin t+C$

$\displaystyle =2\sin\!\left(e^{\sqrt{x}}\right)+C$

$\displaystyle \textbf{Question 46: }~\int \frac{\cos^5 x}{\sin x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\cos^{5}x}{\sin x}\,dx$

$\displaystyle =\int \frac{\cos^{4}x\cdot \cos x}{\sin x}\,dx$

$\displaystyle =\int \frac{(\cos^{2}x)^{2}\cos x}{\sin x}\,dx$

$\displaystyle =\int \frac{(1-\sin^{2}x)^{2}\cos x}{\sin x}\,dx$

$\displaystyle \text{Let } \sin x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\cos x$

$\displaystyle \Rightarrow \cos x\,dx=dt$

$\displaystyle \text{Now, } \int \frac{(1-\sin^{2}x)^{2}\cos x}{\sin x}\,dx=\int \frac{(1-t^{2})^{2}}{t}\,dt$

$\displaystyle =\int \frac{1-2t^{2}+t^{4}}{t}\,dt$

$\displaystyle =\int \left(\frac{1}{t}-2t+t^{3}\right)dt$

$\displaystyle =\log|t|-t^{2}+\frac{t^{4}}{4}+C$

$\displaystyle =\log|\sin x|-\sin^{2}x+\frac{\sin^{4}x}{4}+C$

$\displaystyle \textbf{Question 47: }~\int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx$

$\displaystyle \text{Let } \sqrt{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$

$\displaystyle \text{Now, } \int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx=2\int \sin t\,dt$

$\displaystyle =2(-\cos t)+C$

$\displaystyle =-2\cos \sqrt{x}+C$

$\displaystyle \textbf{Question 48: }~\int \frac{(x+1)e^x}{\sin^2(xe^x)}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{(x+1)e^{x}}{\sin^{2}(x e^{x})}\,dx$

$\displaystyle \text{Let } x e^{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=e^{x}+x e^{x}$

$\displaystyle \Rightarrow (x+1)e^{x}\,dx=dt$

$\displaystyle \text{Now, } \int \frac{(x+1)e^{x}}{\sin^{2}(x e^{x})}\,dx=\int \frac{dt}{\sin^{2}t}$

$\displaystyle =\int \mathrm{cosec}^{2}t\,dt$

$\displaystyle =-\cot t+C$

$\displaystyle =-\cot(x e^{x})+C$

$\displaystyle \textbf{Question 49: }~\int 5^{x+\tan^{-1}x}\left(\frac{x^2+2}{x^2+1}\right)\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int 5^{x+\tan^{-1}x}\left(\frac{x^{2}+2}{x^{2}+1}\right)\,dx$

$\displaystyle \text{Let } x+\tan^{-1}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=1+\frac{1}{1+x^{2}}$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{x^{2}+2}{x^{2}+1}$

$\displaystyle \Rightarrow \left(\frac{x^{2}+2}{x^{2}+1}\right)\,dx=dt$

$\displaystyle \text{Now, } \int 5^{x+\tan^{-1}x}\left(\frac{x^{2}+2}{x^{2}+1}\right)\,dx=\int 5^{t}\,dt$

$\displaystyle =\frac{5^{t}}{\log 5}+C$

$\displaystyle =\frac{5^{x+\tan^{-1}x}}{\log 5}+C$

$\displaystyle \textbf{Question 50: }~\int \frac{e^{m\sin^{-1}x}}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{e^{m\sin^{-1}x}}{\sqrt{1-x^{2}}}\,dx$

$\displaystyle \text{Let } \sin^{-1}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{1-x^{2}}}=dt$

$\displaystyle \text{Now, } \int \frac{e^{m\sin^{-1}x}}{\sqrt{1-x^{2}}}\,dx=\int e^{mt}\,dt$

$\displaystyle =\frac{e^{mt}}{m}+C$

$\displaystyle =\frac{e^{m\sin^{-1}x}}{m}+C$

$\displaystyle \textbf{Question 51: }~\int \frac{\cos \sqrt{x}}{\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\cos \sqrt{x}}{\sqrt{x}}\,dx$

$\displaystyle \text{Let } \sqrt{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$

$\displaystyle \text{Now, } \int \frac{\cos \sqrt{x}}{\sqrt{x}}\,dx=2\int \cos t\,dt$

$\displaystyle =2\sin t+C$

$\displaystyle =2\sin \sqrt{x}+C$

$\displaystyle \textbf{Question 52: }~\int \frac{\sin(\tan^{-1}x)}{1+x^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin(\tan^{-1}x)}{1+x^{2}}\,dx$

$\displaystyle \text{Let } \tan^{-1}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{1+x^{2}}$

$\displaystyle \Rightarrow \frac{dx}{1+x^{2}}=dt$

$\displaystyle \text{Now, } \int \frac{\sin(\tan^{-1}x)}{1+x^{2}}\,dx=\int \sin t\,dt$

$\displaystyle =-\cos t+C$

$\displaystyle =-\cos(\tan^{-1}x)+C$

$\displaystyle \textbf{Question 53: }~\int \frac{\sin(\log x)}{x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin(\log x)}{x}\,dx$

$\displaystyle \text{Let } \log x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{x}$

$\displaystyle \Rightarrow \frac{1}{x}\,dx=dt$

$\displaystyle \text{Now, } \int \frac{\sin(\log x)}{x}\,dx=\int \sin t\,dt$

$\displaystyle =-\cos t+C$

$\displaystyle =-\cos(\log x)+C$

$\displaystyle \textbf{Question 54: }~\int \frac{e^{m\tan^{-1}x}}{1+x^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{e^{m\tan^{-1}x}}{1+x^{2}}\,dx$

$\displaystyle \text{Let } \tan^{-1}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{1+x^{2}}$

$\displaystyle \Rightarrow \frac{dx}{1+x^{2}}=dt$

$\displaystyle \text{Now, } \int \frac{e^{m\tan^{-1}x}}{1+x^{2}}\,dx=\int e^{mt}\,dt$

$\displaystyle =\frac{e^{mt}}{m}+C$

$\displaystyle =\frac{e^{m\tan^{-1}x}}{m}+C$

$\displaystyle \textbf{Question 55: }~\int \frac{x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}}\,dx$

$\displaystyle \text{Let } x^{2}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=2x$

$\displaystyle \Rightarrow x\,dx=\frac{dt}{2}$

$\displaystyle \text{Now, } \int \frac{x}{\sqrt{x^{2}+a^{2}}+\sqrt{x^{2}-a^{2}}}\,dx=\frac{1}{2}\int \frac{dt}{\sqrt{t+a^{2}}+\sqrt{t-a^{2}}}$

$\displaystyle =\frac{1}{2}\int \frac{dt}{\sqrt{t+a^{2}}+\sqrt{t-a^{2}}}\cdot\frac{\sqrt{t+a^{2}}-\sqrt{t-a^{2}}}{\sqrt{t+a^{2}}-\sqrt{t-a^{2}}}$

$\displaystyle =\frac{1}{2}\int \frac{\sqrt{t+a^{2}}-\sqrt{t-a^{2}}}{(t+a^{2})-(t-a^{2})}\,dt$

$\displaystyle =\frac{1}{4a^{2}}\int \left(\sqrt{t+a^{2}}-\sqrt{t-a^{2}}\right)\,dt$

$\displaystyle =\frac{1}{4a^{2}}\left(\int (t+a^{2})^{\frac{1}{2}}\,dt-\int (t-a^{2})^{\frac{1}{2}}\,dt\right)$

$\displaystyle =\frac{1}{4a^{2}}\left(\frac{(t+a^{2})^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(t-a^{2})^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$

$\displaystyle =\frac{1}{6a^{2}}\left((t+a^{2})^{\frac{3}{2}}-(t-a^{2})^{\frac{3}{2}}\right)+C$

$\displaystyle =\frac{1}{6a^{2}}\left((x^{2}+a^{2})^{\frac{3}{2}}-(x^{2}-a^{2})^{\frac{3}{2}}\right)+C$

$\displaystyle \textbf{Question 56: }~\int \frac{x\,\tan^{-1}(x^2)}{1+x^4}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{x\,\tan^{-1}(x^{2})}{1+x^{4}}\,dx$

$\displaystyle \text{Let } \tan^{-1}(x^{2})=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{1+(x^{2})^{2}}\cdot 2x$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{2x}{1+x^{4}}$

$\displaystyle \Rightarrow \frac{x}{1+x^{4}}\,dx=\frac{dt}{2}$

$\displaystyle \text{Now, } \int \frac{x\,\tan^{-1}(x^{2})}{1+x^{4}}\,dx=\frac{1}{2}\int t\,dt$

$\displaystyle =\frac{t^{2}}{4}+C$

$\displaystyle =\frac{\left(\tan^{-1}(x^{2})\right)^{2}}{4}+C$

$\displaystyle \textbf{Question 57: }~\int \frac{(\sin^{-1}x)^3}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{(\sin^{-1}x)^{3}}{\sqrt{1-x^{2}}}\,dx$

$\displaystyle \text{Let } \sin^{-1}x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{1-x^{2}}}=dt$

$\displaystyle \text{Now, } \int \frac{(\sin^{-1}x)^{3}}{\sqrt{1-x^{2}}}\,dx=\int t^{3}\,dt$

$\displaystyle =\frac{t^{4}}{4}+C$

$\displaystyle =\frac{(\sin^{-1}x)^{4}}{4}+C$

$\displaystyle \textbf{Question 58: }~\int \frac{\sin(2+3\log x)}{x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin(2+3\log x)}{x}\,dx$

$\displaystyle \text{Let } 2+3\log x=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{3}{x}$

$\displaystyle \Rightarrow \frac{dx}{x}=\frac{dt}{3}$

$\displaystyle \text{Now, } \int \frac{\sin(2+3\log x)}{x}\,dx=\frac{1}{3}\int \sin t\,dt$

$\displaystyle =-\frac{1}{3}\cos t+C$

$\displaystyle =-\frac{1}{3}\cos(2+3\log x)+C$

$\displaystyle \textbf{Question 59: }~\int x e^{x^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int x\,e^{x^{2}}\,dx$

$\displaystyle \text{Let } x^{2}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=2x$

$\displaystyle \Rightarrow x\,dx=\frac{dt}{2}$

$\displaystyle \text{Now, } \int x\,e^{x^{2}}\,dx=\frac{1}{2}\int e^{t}\,dt$

$\displaystyle =\frac{1}{2}e^{t}+C$

$\displaystyle =\frac{1}{2}e^{x^{2}}+C$

$\displaystyle \textbf{Question 60: }~\int \frac{e^{2x}}{1+e^x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{e^{2x}}{1+e^{x}}\,dx$

$\displaystyle =\int \frac{e^{x}\cdot e^{x}}{1+e^{x}}\,dx$

$\displaystyle \text{Let } 1+e^{x}=t$

$\displaystyle \Rightarrow e^{x}=t-1$

$\displaystyle \Rightarrow e^{x}\,dx=dt$

$\displaystyle \text{Now, } \int \frac{e^{x}\cdot e^{x}}{1+e^{x}}\,dx=\int \frac{t-1}{t}\,dt$

$\displaystyle =\int \left(1-\frac{1}{t}\right)dt$

$\displaystyle =t-\log|t|+C$

$\displaystyle =(1+e^{x})-\log(1+e^{x})+C$

$\displaystyle =e^{x}-\log(1+e^{x})+C$

$\displaystyle \textbf{Question 61: }~\int \frac{\sec^2\sqrt{x}}{\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sec^{2}\!\sqrt{x}}{\sqrt{x}}\,dx$

$\displaystyle \text{Let } \sqrt{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$

$\displaystyle \text{Now, } \int \frac{\sec^{2}\!\sqrt{x}}{\sqrt{x}}\,dx=2\int \sec^{2}t\,dt$

$\displaystyle =2\tan t+C$

$\displaystyle =2\tan(\sqrt{x})+C$

$\displaystyle \textbf{Question 62: }~\int \tan^3 2x~\sec 2x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \tan^{3}(2x)\cdot \sec(2x)\,dx$

$\displaystyle =\int \tan^{2}(2x)\cdot \sec(2x)\tan(2x)\,dx$

$\displaystyle =\int (\sec^{2}(2x)-1)\,\sec(2x)\tan(2x)\,dx$

$\displaystyle \text{Let } \sec(2x)=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=2\sec(2x)\tan(2x)$

$\displaystyle \Rightarrow \sec(2x)\tan(2x)\,dx=\frac{dt}{2}$

$\displaystyle \text{Now, } \int (\sec^{2}(2x)-1)\,\sec(2x)\tan(2x)\,dx=\frac{1}{2}\int (t^{2}-1)\,dt$

$\displaystyle =\frac{1}{2}\left(\frac{t^{3}}{3}-t\right)+C$

$\displaystyle =\frac{t^{3}}{6}-\frac{t}{2}+C$

$\displaystyle =\frac{\sec^{3}(2x)}{6}-\frac{\sec(2x)}{2}+C$

$\displaystyle \textbf{Question 63: }~\int \frac{x+\sqrt{x+1}}{x+2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have, } I=\int \frac{x+\sqrt{x+1}}{x+2}\,dx$

$\displaystyle \text{Let } x+1=t^{2}$

$\displaystyle \Rightarrow \frac{dx}{dt}=2t$

$\displaystyle \Rightarrow dx=2t\,dt$

$\displaystyle \text{Now, } I=\int \frac{t^{2}-1+t}{t^{2}+1}\cdot 2t\,dt$

$\displaystyle =2\int \frac{t^{3}+t^{2}-t}{t^{2}+1}\,dt$

$\displaystyle =2\int \frac{t^{3}+t}{t^{2}+1}\,dt+2\int \frac{t^{2}+1}{t^{2}+1}\,dt-2\int \frac{2t+1}{t^{2}+1}\,dt$

$\displaystyle =2\int t\,dt+2\int dt-2\int \frac{2t}{t^{2}+1}\,dt-2\int \frac{1}{t^{2}+1}\,dt$

$\displaystyle =t^{2}+2t-2\log(t^{2}+1)-2\tan^{-1}t+C$

$\displaystyle =(x+1)+2\sqrt{x+1}-2\log(x+2)-2\tan^{-1}\!\sqrt{x+1}+C$

$\displaystyle \textbf{Question 64: }~\int 5^{5^x}\,5^x\,5^x\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int 5^{5^{x}}\cdot 5^{5^{x}}\cdot 5^{x}\,dx$

$\displaystyle \text{Let } 5^{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=5^{x}\log 5$

$\displaystyle \Rightarrow 5^{x}\,dx=\frac{dt}{\log 5}$

$\displaystyle \text{Now, } \int 5^{5^{x}}\cdot 5^{5^{x}}\cdot 5^{x}\,dx=\int 5^{t}\cdot 5^{t}\cdot \frac{dt}{\log 5}$

$\displaystyle =\frac{1}{\log 5}\int 5^{2t}\,dt$

$\displaystyle \text{Let } 2t=p$

$\displaystyle \Rightarrow dp=2\,dt$

$\displaystyle \Rightarrow dt=\frac{dp}{2}$

$\displaystyle \text{Now, } \frac{1}{\log 5}\int 5^{2t}\,dt=\frac{1}{2\log 5}\int 5^{p}\,dp$

$\displaystyle =\frac{1}{2\log 5}\cdot\frac{5^{p}}{\log 5}+C$

$\displaystyle =\frac{5^{p}}{2(\log 5)^{2}}+C$

$\displaystyle =\frac{5^{2t}}{2(\log 5)^{2}}+C$

$\displaystyle =\frac{5^{2\cdot 5^{x}}}{2(\log 5)^{2}}+C$

$\displaystyle \textbf{Question 65: }~\int \frac{1}{x\sqrt{x^4-1}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{dx}{x\sqrt{x^{4}-1}}$

$\displaystyle =\int \frac{x\,dx}{x^{2}\sqrt{(x^{2})^{2}-1}}$

$\displaystyle \text{Let } x^{2}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=2x$

$\displaystyle \Rightarrow x\,dx=\frac{dt}{2}$

$\displaystyle \text{Now, } \int \frac{x\,dx}{x^{2}\sqrt{(x^{2})^{2}-1}}=\frac{1}{2}\int \frac{dt}{t\sqrt{t^{2}-1}}$

$\displaystyle =\frac{1}{2}\sec^{-1}t+C$

$\displaystyle =\frac{1}{2}\sec^{-1}(x^{2})+C$

$\displaystyle \textbf{Question 66: }~\int \sqrt{e^x-1}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \sqrt{e^{x}-1}\,dx$

$\displaystyle \text{Let } e^{x}-1=t^{2}$

$\displaystyle \Rightarrow e^{x}=t^{2}+1$

$\displaystyle \Rightarrow e^{x}\frac{dt}{dx}=2t$

$\displaystyle \Rightarrow dx=\frac{2t}{e^{x}}\,dt=\frac{2t}{t^{2}+1}\,dt$

$\displaystyle \text{Now, } \int \sqrt{e^{x}-1}\,dx=\int \frac{2t^{2}}{t^{2}+1}\,dt$

$\displaystyle =2\int \frac{t^{2}+1-1}{t^{2}+1}\,dt$

$\displaystyle =2\int dt-2\int \frac{dt}{t^{2}+1}$

$\displaystyle =2t-2\tan^{-1}t+C$

$\displaystyle =2\sqrt{e^{x}-1}-2\tan^{-1}\!\left(\sqrt{e^{x}-1}\right)+C$

$\displaystyle \textbf{Question 67: }~\int \frac{1}{(x+1)(x^2+2x+2)}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } I=\int \frac{dx}{(x+1)(x^{2}+2x+2)}$

$\displaystyle =\int \frac{dx}{(x+1)\left((x+1)^{2}+1\right)}$

$\displaystyle \text{Let } x+1=t$

$\displaystyle \Rightarrow dx=dt$

$\displaystyle \text{Now, } I=\int \frac{dt}{t(t^{2}+1)}$

$\displaystyle =\int \left(\frac{1}{t}-\frac{t}{t^{2}+1}\right)dt$

$\displaystyle =\int \frac{1}{t}\,dt-\int \frac{t}{t^{2}+1}\,dt$

$\displaystyle =\log|t|-\frac{1}{2}\log(t^{2}+1)+C$

$\displaystyle =\log|x+1|-\frac{1}{2}\log\!\left((x+1)^{2}+1\right)+C$

$\displaystyle =\log\!\left(\frac{|x+1|}{\sqrt{x^{2}+2x+2}}\right)+C$

$\displaystyle \textbf{Question 68: }~\int \frac{x^5}{\sqrt{1+x^3}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{x^{5}}{\sqrt{1+x^{3}}}\,dx$

$\displaystyle =\int \frac{x^{3}\cdot x^{2}}{\sqrt{1+x^{3}}}\,dx$

$\displaystyle \text{Let } 1+x^{3}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=3x^{2}$

$\displaystyle \Rightarrow x^{2}\,dx=\frac{dt}{3}$

$\displaystyle \text{Also, } x^{3}=t-1$

$\displaystyle \text{Now, } \int \frac{x^{3}\cdot x^{2}}{\sqrt{1+x^{3}}}\,dx=\frac{1}{3}\int \frac{t-1}{\sqrt{t}}\,dt$

$\displaystyle =\frac{1}{3}\int \left(t^{\frac{1}{2}}-t^{-\frac{1}{2}}\right)dt$

$\displaystyle =\frac{1}{3}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C$

$\displaystyle =\frac{1}{3}\left(\frac{2}{3}t^{\frac{3}{2}}-2t^{\frac{1}{2}}\right)+C$

$\displaystyle =\frac{2}{9}t^{\frac{3}{2}}-\frac{2}{3}t^{\frac{1}{2}}+C$

$\displaystyle =\frac{2}{9}(1+x^{3})^{\frac{3}{2}}-\frac{2}{3}(1+x^{3})^{\frac{1}{2}}+C$

$\displaystyle \textbf{Question 69: }~\int 4x^3\sqrt{5-x^2}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int 4x^{3}\sqrt{5-x^{2}}\,dx$

$\displaystyle =4\int x^{2}\cdot x\sqrt{5-x^{2}}\,dx$

$\displaystyle \text{Let } 5-x^{2}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-2x$

$\displaystyle \Rightarrow x\,dx=-\frac{dt}{2}$

$\displaystyle \text{Also, } x^{2}=5-t$

$\displaystyle \text{Now, } 4\int x^{2}\cdot x\sqrt{5-x^{2}}\,dx=4\int (5-t)\sqrt{t}\left(-\frac{dt}{2}\right)$

$\displaystyle =-2\int (5-t)\,t^{\frac{1}{2}}\,dt$

$\displaystyle =-2\int \left(5t^{\frac{1}{2}}-t^{\frac{3}{2}}\right)dt$

$\displaystyle =-2\left(5\cdot\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right)+C$

$\displaystyle =-2\left(\frac{10}{3}t^{\frac{3}{2}}-\frac{2}{5}t^{\frac{5}{2}}\right)+C$

$\displaystyle =-\frac{20}{3}t^{\frac{3}{2}}+\frac{4}{5}t^{\frac{5}{2}}+C$

$\displaystyle =-\frac{20}{3}(5-x^{2})^{\frac{3}{2}}+\frac{4}{5}(5-x^{2})^{\frac{5}{2}}+C$

$\displaystyle \textbf{Question 70: }~\int \frac{1}{\sqrt{x}+x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{dx}{\sqrt{x}+x}$

$\displaystyle =\int \frac{dx}{\sqrt{x}(1+\sqrt{x})}$

$\displaystyle \text{Let } 1+\sqrt{x}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$

$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$

$\displaystyle \text{Now, } \int \frac{dx}{\sqrt{x}(1+\sqrt{x})}=\int \frac{1}{t}\cdot \frac{dx}{\sqrt{x}}$

$\displaystyle =\int \frac{1}{t}\cdot 2\,dt$

$\displaystyle =2\int \frac{dt}{t}$

$\displaystyle =2\log|t|+C$

$\displaystyle =2\log(1+\sqrt{x})+C$

$\displaystyle \textbf{Question 71: }~\int \frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{dx}{x^{2}(x^{4}+1)^{\frac{3}{4}}}$

$\displaystyle =\int \frac{dx}{x^{2}\left[x^{4}\left(1+\frac{1}{x^{4}}\right)\right]^{\frac{3}{4}}}$

$\displaystyle =\int \frac{dx}{x^{2}\cdot x^{3}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$

$\displaystyle =\int \frac{dx}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$

$\displaystyle =\int \frac{\left(1+\frac{1}{x^{4}}\right)^{\frac{-3}{4}}}{x^{5}}\,dx$

$\displaystyle \text{Let } 1+\frac{1}{x^{4}}=t$

$\displaystyle \Rightarrow \frac{dt}{dx}=-\frac{4}{x^{5}}$

$\displaystyle \Rightarrow \frac{dx}{x^{5}}=-\frac{dt}{4}$

$\displaystyle \text{Now, } \int \frac{dx}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}=\int t^{-\frac{3}{4}}\left(-\frac{dt}{4}\right)$

$\displaystyle =-\frac{1}{4}\int t^{-\frac{3}{4}}\,dt$

$\displaystyle =-\frac{1}{4}\cdot\frac{t^{\frac{1}{4}}}{\frac{1}{4}}+C$

$\displaystyle =-t^{\frac{1}{4}}+C$

$\displaystyle =-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+C$

$\displaystyle \textbf{Question 72: }~\int \frac{\sin^5 x}{\cos^4 x}\,dx.$
$\displaystyle \text{Answer:}$

$\displaystyle \int \frac{\sin^5 x}{\cos^4 x}\,dx$

$\displaystyle = \int \frac{\sin^4 x \sin x}{\cos^4 x}\,dx$

$\displaystyle = \int \frac{(\sin^2 x)^2 \sin x}{\cos^4 x}\,dx$

$\displaystyle = \int \frac{(1-\cos^2 x)^2 \sin x}{\cos^4 x}\,dx$

$\displaystyle = \int \left(\frac{1}{\cos^4 x}-\frac{2}{\cos^2 x}+1\right)\sin x\,dx$

$\displaystyle \text{Let } \cos x = t$

$\displaystyle \Rightarrow -\sin x\,dx = dt$

$\displaystyle \Rightarrow \sin x\,dx = -dt$

$\displaystyle \int \left(\frac{1}{\cos^4 x}-\frac{2}{\cos^2 x}+1\right)\sin x\,dx = -\int (t^{-4}-2t^{-2}+1)\,dt$

$\displaystyle = -\left(\frac{t^{-3}}{-3}-2\frac{t^{-1}}{-1}+t\right)+C$

$\displaystyle = \frac{1}{3t^3}-\frac{2}{t}-t+C$

$\displaystyle = \frac{1}{3\cos^3 x}-\frac{2}{\cos x}-\cos x+C$

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Ultimate Math Study Guide for Students

Class 12: Indefinite Integrals – Exercise 19.9

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