Class 12: Indefinite Integrals

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

$\displaystyle \textbf{Question 1: }~\int x\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\cos x\,dx$
$\displaystyle = x\int \cos x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \cos x\,dx\right)\,dx$
$\displaystyle = x\sin x - \int \sin x\,dx$
$\displaystyle = x\sin x + \cos x + C$

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

$\displaystyle \textbf{Question 3: }~\int x^3\log x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^3\log x\,dx$
$\displaystyle \text{Taking }\log x\text{ as the first function and }x^3\text{ as the second function.}$
$\displaystyle = \log x\int x^3\,dx - \int \left(\frac{d}{dx}\log x\right)\left(\int x^3\,dx\right)\,dx$
$\displaystyle = (\log x)\frac{x^4}{4} - \int \frac{1}{x}\left(\frac{x^4}{4}\right)\,dx$
$\displaystyle = \frac{x^4}{4}\log x - \frac{1}{4}\int x^3\,dx$
$\displaystyle = \frac{x^4}{4}\log x - \frac{x^4}{16} + C$

$\displaystyle \textbf{Question 4: }~\int xe^x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x e^x\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }e^x\text{ as the second function.}$
$\displaystyle = x\int e^x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int e^x\,dx\right)\,dx$
$\displaystyle = x e^x - \int 1\cdot e^x\,dx$
$\displaystyle = x e^x - e^x + C$
$\displaystyle = (x-1)e^x + C$

$\displaystyle \textbf{Question 5: }~\int xe^{2x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x e^{2x}\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }e^{2x}\text{ as the second function.}$
$\displaystyle = x\int e^{2x}\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int e^{2x}\,dx\right)\,dx$
$\displaystyle = x\frac{e^{2x}}{2} - \int \frac{e^{2x}}{2}\,dx$
$\displaystyle = \frac{x}{2}e^{2x} - \frac{e^{2x}}{4} + C$
$\displaystyle = e^{2x}\left(\frac{x}{2} - \frac{1}{4}\right) + C$

$\displaystyle \textbf{Question 6: }~\int x^2e^{-x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^2 e^{-x}\,dx$
$\displaystyle \text{Taking }x^2\text{ as the first function and }e^{-x}\text{ as the second function.}$
$\displaystyle = x^2\int e^{-x}\,dx - \int \left(\frac{d}{dx}(x^2)\right)\left(\int e^{-x}\,dx\right)\,dx$
$\displaystyle = -x^2 e^{-x} - \int 2x(-e^{-x})\,dx$
$\displaystyle = -x^2 e^{-x} + 2\int x e^{-x}\,dx$
$\displaystyle = -x^2 e^{-x} + 2\left[-x e^{-x} + \int e^{-x}\,dx\right]$
$\displaystyle = -x^2 e^{-x} + 2[-x e^{-x} - e^{-x}] + C$
$\displaystyle = -e^{-x}(x^2 + 2x + 2) + C$

$\displaystyle \textbf{Question 7: }~\int x^2\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^2\cos x\,dx$
$\displaystyle \text{Taking }x^2\text{ as the first function and }\cos x\text{ as the second function.}$
$\displaystyle = x^2\int \cos x\,dx - \int \left(\frac{d}{dx}(x^2)\right)\left(\int \cos x\,dx\right)\,dx$
$\displaystyle = x^2\sin x - \int 2x\sin x\,dx$
$\displaystyle = x^2\sin x - 2\left[x\int \sin x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \sin x\,dx\right)\,dx\right]$
$\displaystyle = x^2\sin x - 2\left[-x\cos x + \int \cos x\,dx\right]$
$\displaystyle = x^2\sin x + 2x\cos x - 2\sin x + C$

$\displaystyle \textbf{Question 8: }~\int x^2\cos 2x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^2\cos 2x\,dx$
$\displaystyle \text{Taking }x^2\text{ as the first function and }\cos 2x\text{ as the second function.}$
$\displaystyle = x^2\int \cos 2x\,dx - \int \left(\frac{d}{dx}(x^2)\right)\left(\int \cos 2x\,dx\right)\,dx$
$\displaystyle = \frac{x^2}{2}\sin 2x - \int \frac{2x\sin 2x}{2}\,dx$
$\displaystyle = \frac{x^2}{2}\sin 2x - \int x\sin 2x\,dx$
$\displaystyle = \frac{x^2}{2}\sin 2x - \left[x\int \sin 2x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \sin 2x\,dx\right)\,dx\right]$
$\displaystyle = \frac{x^2}{2}\sin 2x - \left[-\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2}\,dx\right]$
$\displaystyle = \frac{x^2}{2}\sin 2x + \frac{x\cos 2x}{2} - \frac{\sin 2x}{4} + C$

$\displaystyle \textbf{Question 9: }~\int x\sin 2x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\sin 2x\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }\sin 2x\text{ as the second function.}$
$\displaystyle = x\int \sin 2x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \sin 2x\,dx\right)\,dx$
$\displaystyle = x\left(-\frac{\cos 2x}{2}\right) - \int 1\left(-\frac{\cos 2x}{2}\right)\,dx$
$\displaystyle = -\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2}\,dx$
$\displaystyle = -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C$

$\displaystyle \textbf{Question 10: }~\int \frac{\log(\log x)}{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{\log(\log x)}{x}\,dx$
$\displaystyle \text{Taking }\log(\log x)\text{ as the first function and }\frac{1}{x}\text{ as the second function.}$
$\displaystyle = \log(\log x)\int \frac{1}{x}\,dx - \int \left(\frac{d}{dx}\{\log(\log x)\}\right)\left(\int \frac{1}{x}\,dx\right)\,dx$
$\displaystyle = \log x\cdot \log(\log x) - \int \frac{1}{x\log x}\cdot \log x\,dx$
$\displaystyle = \log x\cdot \log(\log x) - \int \frac{1}{x}\,dx$
$\displaystyle = \log x\cdot \log(\log x) - \log x + C$

$\displaystyle \textbf{Question 11: }~\int x^2\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^2\cos x\,dx$
$\displaystyle \text{Taking }x^2\text{ as the first function and }\cos x\text{ as the second function.}$
$\displaystyle = x^2\int \cos x\,dx - \int \left(\frac{d}{dx}(x^2)\right)\left(\int \cos x\,dx\right)\,dx$
$\displaystyle = x^2\sin x - \int 2x\sin x\,dx$
$\displaystyle = x^2\sin x - 2\left[x\int \sin x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \sin x\,dx\right)\,dx\right]$
$\displaystyle = x^2\sin x - 2\left[x(-\cos x) - \int 1(-\cos x)\,dx\right]$
$\displaystyle = x^2\sin x - 2\left[-x\cos x + \int \cos x\,dx\right]$
$\displaystyle = x^2\sin x - 2\left[-x\cos x + \sin x\right]$
$\displaystyle = x^2\sin x + 2x\cos x - 2\sin x + C$

$\displaystyle \textbf{Question 12: }~\int x\mathrm{cosec}^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\,\mathrm{cosec}^2 x\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }\mathrm{cosec}^2 x\text{ as the second function.}$
$\displaystyle = x\int \mathrm{cosec}^2 x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \mathrm{cosec}^2 x\,dx\right)\,dx$
$\displaystyle = x(-\cot x) - \int 1(-\cot x)\,dx$
$\displaystyle = -x\cot x + \int \cot x\,dx$
$\displaystyle = -x\cot x + \log|\sin x| + C$

$\displaystyle \textbf{Question 13: }~\int x\cos^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\cos^2 x\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }\cos^2 x\text{ as the second function.}$
$\displaystyle = x\int \frac{1+\cos 2x}{2}\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \frac{1+\cos 2x}{2}\,dx\right)\,dx$
$\displaystyle = x\left(\frac{x}{2}+\frac{\sin 2x}{4}\right) - \int \frac{1}{2}\left(x+\frac{\sin 2x}{2}\right)\,dx$
$\displaystyle = \frac{x^2}{2}+\frac{x\sin 2x}{4} - \left(\frac{x^2}{4}-\frac{\cos 2x}{8}\right) + C$
$\displaystyle = \frac{x^2}{4}+\frac{x\sin 2x}{4}+\frac{\cos 2x}{8}+C$

$\displaystyle \textbf{Question 14: }~\int x^n\log x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^n\log x\,dx$
$\displaystyle = \int (\log x)\,x^n\,dx$
$\displaystyle \int u\,v\,dx = u\int v\,dx - \int \left(\frac{du}{dx}\right)\left(\int v\,dx\right)\,dx$
$\displaystyle = \log x\int x^n\,dx - \int \left(\frac{d}{dx}(\log x)\right)\left(\int x^n\,dx\right)\,dx$
$\displaystyle = \log x\cdot \frac{x^{n+1}}{n+1} - \int \frac{1}{x}\cdot \frac{x^{n+1}}{n+1}\,dx$
$\displaystyle = \frac{x^{n+1}\log x}{n+1} - \frac{1}{n+1}\int x^n\,dx$
$\displaystyle = \frac{x^{n+1}\log x}{n+1} - \frac{x^{n+1}}{(n+1)^2} + C$
$\displaystyle = \frac{x^{n+1}}{n+1}\left(\log x - \frac{1}{n+1}\right) + C$

$\displaystyle \textbf{Question 15: }~\int \frac{\log x}{x^n}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{1}{x^n}\log x\,dx$
$\displaystyle \text{Taking }\log x\text{ as the first function and }\frac{1}{x^n}\text{ as the second function.}$
$\displaystyle = \log x\int \frac{1}{x^n}\,dx - \int \left(\frac{d}{dx}(\log x)\right)\left(\int \frac{1}{x^n}\,dx\right)\,dx$
$\displaystyle = \log x\left(\frac{x^{1-n}}{1-n}\right) - \int \frac{1}{x}\left(\frac{x^{1-n}}{1-n}\right)\,dx$
$\displaystyle = \log x\left(\frac{x^{1-n}}{1-n}\right) - \frac{1}{1-n}\int x^{-n}\,dx$
$\displaystyle = \log x\left(\frac{x^{1-n}}{1-n}\right) - \frac{x^{1-n}}{(1-n)^2} + C$

$\displaystyle \textbf{Question 16: }~\int x^2\sin^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^2\sin^2 x\,dx$
$\displaystyle \text{Taking }x^2\text{ as the first function and }\sin^2 x\text{ as the second function.}$
$\displaystyle = x^2\int \frac{1-\cos 2x}{2}\,dx - \int \left(\frac{d}{dx}(x^2)\right)\left(\int \frac{1-\cos 2x}{2}\,dx\right)\,dx$
$\displaystyle = \frac{x^2}{2}\left(x-\frac{\sin 2x}{2}\right) - \int 2x\left(\frac{x}{2}-\frac{\sin 2x}{4}\right)\,dx$
$\displaystyle = \frac{x^3}{2}-\frac{x^2\sin 2x}{4} - \int x^2\,dx + \int \frac{x\sin 2x}{2}\,dx$
$\displaystyle \text{Here, taking }x\text{ as the first function and }\sin 2x\text{ as the second function.}$
$\displaystyle = \frac{x^3}{2}-\frac{x^2\sin 2x}{4}-\frac{x^3}{3}+\frac{1}{2}\left[x\int \sin 2x\,dx-\int \left(\frac{d}{dx}(x)\right)\left(\int \sin 2x\,dx\right)\,dx\right]$
$\displaystyle = \frac{x^3}{2}-\frac{x^2\sin 2x}{4}-\frac{x^3}{3}+\frac{1}{2}\left[-\frac{x\cos 2x}{2}+\int \frac{\cos 2x}{2}\,dx\right]$
$\displaystyle = \frac{x^3}{6}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C$

$\displaystyle \textbf{Question 17: }~\int 2x^3e^{x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int 2x^3 e^{x^2}\,dx$
$\displaystyle = \int x^2\,(e^{x^2})\cdot 2x\,dx$
$\displaystyle \text{Let }x^2=t$
$\displaystyle \Rightarrow 2x\,dx=dt$
$\displaystyle = \int t\,e^t\,dt$
$\displaystyle = t\int e^t\,dt - \int \left(\frac{d}{dt}(t)\right)\left(\int e^t\,dt\right)\,dt$
$\displaystyle = t e^t - \int e^t\,dt$
$\displaystyle = t e^t - e^t + C$
$\displaystyle = x^2 e^{x^2} - e^{x^2} + C$
$\displaystyle = e^{x^2}(x^2-1) + C$

$\displaystyle \textbf{Question 18: }~\int x^3\cos x^2\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^3\cos x^2\,dx$
$\displaystyle \text{Let }x^2=t$
$\displaystyle \Rightarrow \frac{dt}{dx}=2x$
$\displaystyle \Rightarrow dx=\frac{dt}{2x}$
$\displaystyle = \frac{1}{2}\int t\cos t\,dt$
$\displaystyle \text{Taking }t\text{ as the first function and }\cos t\text{ as the second function.}$
$\displaystyle = \frac{1}{2}\left[t\int \cos t\,dt - \int \left(\frac{d}{dt}(t)\right)\left(\int \cos t\,dt\right)\,dt\right]$
$\displaystyle = \frac{1}{2}\left[t\sin t - \int \sin t\,dt\right]$
$\displaystyle = \frac{1}{2}\left[t\sin t + \cos t\right] + C$
$\displaystyle = \frac{x^2\sin x^2}{2} + \frac{\cos x^2}{2} + C$

$\displaystyle \textbf{Question 19: }~\int x\sin x\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\sin x\cdot \cos x\,dx$
$\displaystyle = \frac{1}{2}\int x(2\sin x\cos x)\,dx$
$\displaystyle = \frac{1}{2}\int x\sin 2x\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }\sin 2x\text{ as the second function.}$
$\displaystyle = \frac{1}{2}\left[x\int \sin 2x\,dx - \int \left(\frac{d}{dx}(x)\right)\left(\int \sin 2x\,dx\right)\,dx\right]$
$\displaystyle = \frac{1}{2}\left[x\left(-\frac{\cos 2x}{2}\right) - \int 1\left(-\frac{\cos 2x}{2}\right)\,dx\right]$
$\displaystyle = \frac{1}{2}\left[-\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2}\,dx\right]$
$\displaystyle = \frac{1}{2}\left[-\frac{x\cos 2x}{2} + \frac{\sin 2x}{4}\right] + C$
$\displaystyle = -\frac{x\cos 2x}{4} + \frac{\sin 2x}{8} + C$

$\displaystyle \textbf{Question 20: }~\int \sin x\log(\cos x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \sin x\cdot \log(\cos x)\,dx$
$\displaystyle \text{Let }\cos x=t$
$\displaystyle \Rightarrow \frac{dt}{dx}=-\sin x$
$\displaystyle \Rightarrow \sin x\,dx=-dt$
$\displaystyle \therefore I=-\int \log t\,dt$
$\displaystyle =-\int 1\cdot \log t\,dt$
$\displaystyle \text{Taking }\log t\text{ as the first function and }1\text{ as the second function.}$
$\displaystyle =-\left[\log t\int 1\,dt-\int \left(\frac{d}{dt}(\log t)\right)\left(\int 1\,dt\right)\,dt\right]$
$\displaystyle =-\left[\log t\cdot t-\int \frac{1}{t}\cdot t\,dt\right]$
$\displaystyle =-\left[\log t\cdot t-\int 1\,dt\right]$
$\displaystyle =-[t\log t-t]+C$
$\displaystyle =-t(\log t-1)+C$
$\displaystyle \text{Substituting }t=\cos x$
$\displaystyle =-\cos x\{\log(\cos x)-1\}+C$
$\displaystyle =\cos x\{1-\log(\cos x)\}+C$

$\displaystyle \textbf{Question 21: }~\int (\log x)^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int (\log x)^2\,x\,dx$
$\displaystyle \text{Taking }(\log x)^2\text{ as the first function and }x\text{ as the second function.}$
$\displaystyle = (\log x)^2\int x\,dx - \int \left(\frac{d}{dx}(\log x)^2\right)\left(\int x\,dx\right)\,dx$
$\displaystyle = (\log x)^2\cdot \frac{x^2}{2} - \int \frac{2\log x}{x}\cdot \frac{x^2}{2}\,dx$
$\displaystyle = (\log x)^2\cdot \frac{x^2}{2} - \int x\log x\,dx$
$\displaystyle = (\log x)^2\cdot \frac{x^2}{2} - \left[\log x\int x\,dx - \int \left(\frac{d}{dx}(\log x)\right)\left(\int x\,dx\right)\,dx\right]$
$\displaystyle = (\log x)^2\cdot \frac{x^2}{2} - \left[\log x\cdot \frac{x^2}{2} - \int \frac{1}{x}\cdot \frac{x^2}{2}\,dx\right]$
$\displaystyle = (\log x)^2\cdot \frac{x^2}{2} - \log x\cdot \frac{x^2}{2} + \frac{x^2}{4} + C$
$\displaystyle = \frac{x^2}{2}\left[(\log x)^2 - \log x + \frac{1}{2}\right] + C$

$\displaystyle \textbf{Question 22: }~\int e^{\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int e^{\sqrt{x}}\,dx$
$\displaystyle = \int \sqrt{x}\cdot \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx$
$\displaystyle \text{Let }\sqrt{x}=t$
$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$
$\displaystyle \Rightarrow \frac{1}{2\sqrt{x}}\,dx=dt$
$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$
$\displaystyle \therefore I=2\int t e^t\,dt$
$\displaystyle \text{Taking }t\text{ as the first function and }e^t\text{ as the second function.}$
$\displaystyle =2\left[t\int e^t\,dt-\int \left(\frac{d}{dt}(t)\right)\left(\int e^t\,dt\right)\,dt\right]$
$\displaystyle =2\left[t e^t-\int 1\cdot e^t\,dt\right]$
$\displaystyle =2\left[t e^t-e^t\right]+C$
$\displaystyle =2e^t(t-1)+C$
$\displaystyle \text{Substituting }t=\sqrt{x}$
$\displaystyle =2e^{\sqrt{x}}(\sqrt{x}-1)+C$

$\displaystyle \textbf{Question 23: }~\int \frac{\log(x+2)}{(x+2)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \frac{\log(x+2)}{(x+2)^2}\,dx$
$\displaystyle \text{Let }\log(x+2)=t$
$\displaystyle \Rightarrow x+2=e^t$
$\displaystyle \Rightarrow \frac{1}{x+2}\,dx=dt$
$\displaystyle \Rightarrow dx=e^t\,dt$
$\displaystyle \therefore I=\int \frac{t}{(e^t)^2}\cdot e^t\,dt$
$\displaystyle = \int t e^{-t}\,dt$
$\displaystyle \text{Taking }t\text{ as the first function and }e^{-t}\text{ as the second function.}$
$\displaystyle = t\int e^{-t}\,dt - \int \left(\frac{d}{dt}(t)\right)\left(\int e^{-t}\,dt\right)\,dt$
$\displaystyle = t\left(\frac{e^{-t}}{-1}\right) - \int 1\left(\frac{e^{-t}}{-1}\right)\,dt$
$\displaystyle = -t e^{-t} + \int e^{-t}\,dt$
$\displaystyle = -t e^{-t} - e^{-t} + C$
$\displaystyle = -e^{-t}(t+1)+C$
$\displaystyle = -\frac{t+1}{e^t}+C$
$\displaystyle \text{Substituting }t=\log(x+2)\text{ and }e^t=x+2$
$\displaystyle = -\frac{\log(x+2)+1}{x+2}+C$
$\displaystyle = -\frac{\log(x+2)}{x+2}-\frac{1}{x+2}+C$

$\displaystyle \textbf{Question 24: }~\int \frac{x+\sin x}{1+\cos x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x+\sin x}{1+\cos x}\,dx$
$\displaystyle = \int \left[\frac{x}{1+\cos x}+\frac{\sin x}{1+\cos x}\right]dx$
$\displaystyle = \int \left[\frac{x}{2\cos^2\left(\frac{x}{2}\right)}+\frac{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)}\right]dx$
$\displaystyle = \frac{1}{2}\int x\,\sec^2\left(\frac{x}{2}\right)dx+\int \tan\left(\frac{x}{2}\right)dx$
$\displaystyle \text{Let }t=\frac{x}{2}\Rightarrow x=2t,\;dx=2dt$
$\displaystyle \frac{1}{2}\int x\,\sec^2\left(\frac{x}{2}\right)dx=\frac{1}{2}\int (2t)\sec^2(t)\cdot 2dt=2\int t\sec^2 t\,dt$
$\displaystyle \text{Taking }t\text{ as the first function and }\sec^2 t\text{ as the second function.}$
$\displaystyle 2\int t\sec^2 t\,dt=2\left[t\int \sec^2 t\,dt-\int \left(\frac{d}{dt}(t)\right)\left(\int \sec^2 t\,dt\right)dt\right]$
$\displaystyle =2\left[t\tan t-\int \tan t\,dt\right]$
$\displaystyle =2\left[t\tan t-\log|\sec t|\right]+C$
$\displaystyle =x\tan\left(\frac{x}{2}\right)-2\log\left|\sec\left(\frac{x}{2}\right)\right|+C$
$\displaystyle \int \tan\left(\frac{x}{2}\right)dx=2\int \tan t\,dt=2\log|\sec t|+C=2\log\left|\sec\left(\frac{x}{2}\right)\right|+C$
$\displaystyle \therefore \int \frac{x+\sin x}{1+\cos x}\,dx=x\tan\left(\frac{x}{2}\right)+C$

$\displaystyle \textbf{Question 25: }~\int \log_{10}x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \log_{10} x\,dx$
$\displaystyle = \int \frac{\log x}{\log 10}\,dx$
$\displaystyle = \frac{1}{\log 10}\int 1\cdot \log x\,dx$
$\displaystyle \text{Taking }\log x\text{ as the first function and }1\text{ as the second function.}$
$\displaystyle = \frac{1}{\log 10}\left[\log x\int 1\,dx-\int \left(\frac{d}{dx}(\log x)\right)\left(\int 1\,dx\right)\,dx\right]$
$\displaystyle = \frac{1}{\log 10}\left[\log x\cdot x-\int \frac{1}{x}\cdot x\,dx\right]$
$\displaystyle = \frac{1}{\log 10}\left[x\log x-\int 1\,dx\right]+C$
$\displaystyle = \frac{1}{\log 10}\left[x\log x-x\right]+C$
$\displaystyle = \frac{x(\log x-1)}{\log 10}+C$

$\displaystyle \textbf{Question 26: }~\int \cos\sqrt{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \cos\sqrt{x}\,dx$
$\displaystyle = \int \frac{\sqrt{x}\cdot \cos\sqrt{x}}{\sqrt{x}}\,dx$
$\displaystyle \text{Let }\sqrt{x}=t$
$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$
$\displaystyle \Rightarrow \frac{1}{2\sqrt{x}}\,dx=dt$
$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$
$\displaystyle \therefore I=2\int t\cos t\,dt$
$\displaystyle \text{Taking }t\text{ as the first function and }\cos t\text{ as the second function.}$
$\displaystyle =2\left[t\int \cos t\,dt-\int \left(\frac{d}{dt}(t)\right)\left(\int \cos t\,dt\right)\,dt\right]$
$\displaystyle =2\left[t\sin t-\int \sin t\,dt\right]$
$\displaystyle =2\left[t\sin t+\cos t\right]+C$
$\displaystyle \text{Substituting }t=\sqrt{x}$
$\displaystyle =2\left[\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x}\right]+C$

$\displaystyle \textbf{Question 27: }~\int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx$
$\displaystyle \text{Let the first function be }\cos^{-1}x\text{ and the second function be }\frac{x}{\sqrt{1-x^2}}$
$\displaystyle \text{First, we find the integral of the second function, i.e., }\int \frac{x}{\sqrt{1-x^2}}\,dx$
$\displaystyle \text{Put }t=1-x^2$
$\displaystyle \Rightarrow dt=-2x\,dx$
$\displaystyle \int \frac{x}{\sqrt{1-x^2}}\,dx=-\frac{1}{2}\int \frac{1}{\sqrt{t}}\,dt$
$\displaystyle =-\sqrt{t}$
$\displaystyle =-\sqrt{1-x^2}$
$\displaystyle \text{Hence, using integration by parts, we get}$
$\displaystyle \int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx=\cos^{-1}x\int \frac{x}{\sqrt{1-x^2}}\,dx-\int \left(\frac{d}{dx}(\cos^{-1}x)\right)\left(\int \frac{x}{\sqrt{1-x^2}}\,dx\right)\,dx$
$\displaystyle =(\cos^{-1}x)(-\sqrt{1-x^2})-\int \left(-\frac{1}{\sqrt{1-x^2}}\right)(-\sqrt{1-x^2})\,dx$
$\displaystyle =-\sqrt{1-x^2}\cos^{-1}x-\int 1\,dx$
$\displaystyle =-\sqrt{1-x^2}\cos^{-1}x-x+C$

$\displaystyle \textbf{Question 28: }~\int \frac{\log x}{(x+1)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \frac{\log x}{(x+1)^2}\,dx$
$\displaystyle \text{Let the first function be }\log x\text{ and the second function be }\frac{1}{(x+1)^2}$
$\displaystyle \text{First, we find the integral of the second function, i.e., }\int \frac{1}{(x+1)^2}\,dx$
$\displaystyle \text{Put }t=x+1$
$\displaystyle \Rightarrow dt=dx$
$\displaystyle \int \frac{1}{(x+1)^2}\,dx=\int t^{-2}\,dt$
$\displaystyle =-\frac{1}{t}$
$\displaystyle =-\frac{1}{x+1}$
$\displaystyle \text{Hence, using integration by parts, we get}$
$\displaystyle \int \frac{\log x}{(x+1)^2}\,dx=\log x\int \frac{1}{(x+1)^2}\,dx-\int \left(\frac{d}{dx}(\log x)\right)\left(\int \frac{1}{(x+1)^2}\,dx\right)\,dx$
$\displaystyle =(\log x)\left(-\frac{1}{x+1}\right)-\int \left(\frac{1}{x}\right)\left(-\frac{1}{x+1}\right)\,dx$
$\displaystyle =-\frac{\log x}{x+1}+\int \frac{1}{x(x+1)}\,dx$
$\displaystyle =-\frac{\log x}{x+1}+\int \left(\frac{1}{x}-\frac{1}{x+1}\right)\,dx$
$\displaystyle =-\frac{\log x}{x+1}+\log|x|-\log|x+1|+C$
$\displaystyle =-\frac{\log x}{x+1}+\log\left|\frac{x}{x+1}\right|+C$

$\displaystyle \textbf{Question 29: }~\int \mathrm{cosec}^3 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \mathrm{cosec}^3 x\,dx$
$\displaystyle =\int \mathrm{cosec}^2 x\cdot \mathrm{cosec} x\,dx$
$\displaystyle =\int \mathrm{cosec}^2 x\cdot \sqrt{1+\cot^2 x}\,dx$
$\displaystyle \text{Let }\cot x=t$
$\displaystyle \Rightarrow \frac{dt}{dx}=-\mathrm{cosec}^2 x$
$\displaystyle \Rightarrow -\mathrm{cosec}^2 x\,dx=dt$
$\displaystyle \therefore I=-\int \sqrt{1+t^2}\,dt$
$\displaystyle =-\left[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}\log\left|t+\sqrt{1+t^2}\right|\right]+C$
$\displaystyle \text{Substituting }t=\cot x$
$\displaystyle =-\frac{1}{2}\cot x\sqrt{1+\cot^2 x}-\frac{1}{2}\log\left|\cot x+\sqrt{1+\cot^2 x}\right|+C$
$\displaystyle =-\frac{1}{2}\cot x\,\mathrm{cosec} x-\frac{1}{2}\log\left|\cot x+\mathrm{cosec} x\right|+C$
$\displaystyle =-\frac{1}{2}\mathrm{cosec} x\cot x+\frac{1}{2}\log\left|\tan\frac{x}{2}\right|+C$

$\displaystyle \textbf{Question 30: }~\int \sec^{-1}\sqrt{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \mathrm{sec}^{-1}\sqrt{x}\,dx$
$\displaystyle \text{Let } \sqrt{x}=t$
$\displaystyle \Rightarrow x=t^2$
$\displaystyle \Rightarrow \frac{dx}{dt}=2t$
$\displaystyle \Rightarrow dx=2t\,dt$
$\displaystyle \therefore I=\int \mathrm{sec}^{-1} t\cdot 2t\,dt$
$\displaystyle =2\int t\,\mathrm{sec}^{-1} t\,dt$
$\displaystyle \text{Using integration by parts}$
$\displaystyle \text{Let } u=\mathrm{sec}^{-1} t,\quad dv=t\,dt$
$\displaystyle \Rightarrow du=\frac{1}{t\sqrt{t^2-1}}\,dt,\quad v=\frac{t^2}{2}$
$\displaystyle \therefore I=2\left[\frac{t^2}{2}\mathrm{sec}^{-1} t-\int \frac{t^2}{2}\cdot\frac{1}{t\sqrt{t^2-1}}\,dt\right]$
$\displaystyle =t^2\mathrm{sec}^{-1} t-\int \frac{t}{\sqrt{t^2-1}}\,dt$
$\displaystyle =t^2\mathrm{sec}^{-1} t-\sqrt{t^2-1}+C$
$\displaystyle \text{Substituting } t=\sqrt{x}$
$\displaystyle =x\,\mathrm{sec}^{-1}\sqrt{x}-\sqrt{x-1}+C$

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

$\displaystyle \textbf{Question 31: }~\int \sin^{-1}\!\sqrt{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \sin^{-1}\sqrt{x}\,dx$
$\displaystyle = \int \frac{\sqrt{x}\cdot \sin^{-1}\sqrt{x}}{\sqrt{x}}\,dx$
$\displaystyle \text{Let }\sqrt{x}=t$
$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$
$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$
$\displaystyle \therefore I=2\int t\sin^{-1}t\,dt$
$\displaystyle \text{Taking }\sin^{-1}t\text{ as the first function and }t\text{ as the second function.}$
$\displaystyle =2\left[\sin^{-1}t\int t\,dt-\int \left(\frac{d}{dt}(\sin^{-1}t)\right)\left(\int t\,dt\right)\,dt\right]$
$\displaystyle =2\left[\sin^{-1}t\cdot \frac{t^2}{2}-\int \frac{1}{\sqrt{1-t^2}}\cdot \frac{t^2}{2}\,dt\right]$
$\displaystyle =\sin^{-1}t\cdot t^2-\int \frac{t^2}{\sqrt{1-t^2}}\,dt$
$\displaystyle =\sin^{-1}t\cdot t^2-\int \frac{(1-t^2)-1}{\sqrt{1-t^2}}\,dt$
$\displaystyle =\sin^{-1}t\cdot t^2+\int \sqrt{1-t^2}\,dt-\int \frac{1}{\sqrt{1-t^2}}\,dt$
$\displaystyle =\sin^{-1}t\cdot t^2+\frac{t}{2}\sqrt{1-t^2}-\frac{1}{2}\sin^{-1}t+C$
$\displaystyle \text{Substituting }t=\sqrt{x}$
$\displaystyle =x\sin^{-1}\sqrt{x}+\frac{\sqrt{x}}{2}\sqrt{1-x}-\frac{1}{2}\sin^{-1}\sqrt{x}+C$
$\displaystyle =\frac{2x-1}{2}\sin^{-1}\sqrt{x}+\frac{\sqrt{x-x^2}}{2}+C$

$\displaystyle \textbf{Question 32: }~\int x\tan^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\tan^2 x\,dx$
$\displaystyle = \int x(\sec^2 x-1)\,dx$
$\displaystyle = \int x\sec^2 x\,dx-\int x\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }\sec^2 x\text{ as the second function.}$
$\displaystyle = x\int \sec^2 x\,dx-\int \left(\frac{d}{dx}(x)\right)\left(\int \sec^2 x\,dx\right)\,dx-\frac{x^2}{2}$
$\displaystyle = x\tan x-\int 1\cdot \tan x\,dx-\frac{x^2}{2}$
$\displaystyle = x\tan x-\log|\sec x|-\frac{x^2}{2}+C$

$\displaystyle \textbf{Question 33: }~\int x\!\left(\frac{\sec 2x-1}{\sec 2x+1}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\left(\frac{\sec 2x-1}{\sec 2x+1}\right)dx$
$\displaystyle = \int x\left(\frac{\frac{1}{\cos 2x}-1}{\frac{1}{\cos 2x}+1}\right)dx$
$\displaystyle = \int x\left(\frac{1-\cos 2x}{1+\cos 2x}\right)dx$
$\displaystyle = \int x\left(\frac{2\sin^2 x}{2\cos^2 x}\right)dx$
$\displaystyle = \int x\tan^2 x\,dx$
$\displaystyle = \int x(\sec^2 x-1)\,dx$
$\displaystyle = \int x\sec^2 x\,dx-\int x\,dx$
$\displaystyle \text{Taking }x\text{ as the first function and }\sec^2 x\text{ as the second function.}$
$\displaystyle = x\int \sec^2 x\,dx-\int \left(\frac{d}{dx}(x)\right)\left(\int \sec^2 x\,dx\right)\,dx-\frac{x^2}{2}$
$\displaystyle = x\tan x-\int 1\cdot \tan x\,dx-\frac{x^2}{2}$
$\displaystyle = x\tan x-\log|\sec x|-\frac{x^2}{2}+C$

$\displaystyle \textbf{Question 34: }~\int (x+1)e^x\log(xe^x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int (x+1)e^x\log(xe^x)\,dx$
$\displaystyle \text{Let }t=xe^x$
$\displaystyle \Rightarrow \frac{dt}{dx}=(x+1)e^x$
$\displaystyle \Rightarrow (x+1)e^x\,dx=dt$
$\displaystyle \therefore \int (x+1)e^x\log(xe^x)\,dx=\int 1\cdot \log t\,dt$
$\displaystyle \text{Taking }\log t\text{ as the first function and }1\text{ as the second function.}$
$\displaystyle = \log t\int 1\,dt-\int \left(\frac{d}{dt}(\log t)\right)\left(\int 1\,dt\right)\,dt$
$\displaystyle = \log t\cdot t-\int \frac{1}{t}\cdot t\,dt$
$\displaystyle = t\log t-\int 1\,dt$
$\displaystyle = t\log t-t+C$
$\displaystyle \text{Substituting }t=xe^x$
$\displaystyle = (xe^x)\log(xe^x)-xe^x+C$
$\displaystyle = xe^x\{\log(xe^x)-1\}+C$

$\displaystyle \textbf{Question 35: }~\int \sin^{-1}(3x-4x^3)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sin^{-1}(3x-4x^3)\,dx$
$\displaystyle \text{Let }x=\sin\theta$
$\displaystyle \Rightarrow dx=\cos\theta\,d\theta$
$\displaystyle \text{and }\theta=\sin^{-1}x$
$\displaystyle \int \sin^{-1}(3x-4x^3)\,dx=\int \sin^{-1}(3\sin\theta-4\sin^3\theta)\cos\theta\,d\theta$
$\displaystyle =\int \sin^{-1}(\sin 3\theta)\cos\theta\,d\theta$
$\displaystyle =\int 3\theta\cos\theta\,d\theta$
$\displaystyle =3\int \theta\cos\theta\,d\theta$
$\displaystyle =3\left[\theta\int \cos\theta\,d\theta-\int \left(\frac{d}{d\theta}(\theta)\right)\left(\int \cos\theta\,d\theta\right)d\theta\right]$
$\displaystyle =3\left[\theta\sin\theta-\int 1\cdot \sin\theta\,d\theta\right]$
$\displaystyle =3\left[\theta\sin\theta+\cos\theta\right]+C$
$\displaystyle =3\left[\theta\sin\theta+\sqrt{1-\sin^2\theta}\right]+C$
$\displaystyle =3\left[x\sin^{-1}x+\sqrt{1-x^2}\right]+C$

$\displaystyle \textbf{Question 36: }~\int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\,dx$
$\displaystyle \text{Let }x=\tan\theta$
$\displaystyle \Rightarrow dx=\sec^2\theta\,d\theta$
$\displaystyle \therefore \int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\,dx=\int \sin^{-1}\!\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)\sec^2\theta\,d\theta$
$\displaystyle =\int \sin^{-1}(\sin 2\theta)\sec^2\theta\,d\theta$
$\displaystyle =\int 2\theta\,\sec^2\theta\,d\theta$
$\displaystyle =2\int \theta\,\sec^2\theta\,d\theta$
$\displaystyle =2\left[\theta\int \sec^2\theta\,d\theta-\int \left(\frac{d}{d\theta}(\theta)\right)\left(\int \sec^2\theta\,d\theta\right)d\theta\right]$
$\displaystyle =2\left[\theta\tan\theta-\int \tan\theta\,d\theta\right]$
$\displaystyle =2\left[\theta\tan\theta+\log(\cos\theta)\right]+C$
$\displaystyle =2\left[\theta\tan\theta-\frac{1}{2}\log(1+\tan^2\theta)\right]+C$
$\displaystyle \text{Substituting }\theta=\tan^{-1}x$
$\displaystyle =2x\tan^{-1}x-\log|1+x^2|+C$

$\displaystyle \textbf{Question 37: }~\int \tan^{-1}\!\left(\frac{3x-x^3}{1-3x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \tan^{-1}\!\left(\frac{3x-x^3}{1-3x^2}\right)\,dx$
$\displaystyle =\int 3\tan^{-1}(x)\,dx$
$\displaystyle =3\int \tan^{-1}(x)\cdot 1\,dx$
$\displaystyle =3\left[\tan^{-1}x\int 1\,dx-\int \left(\frac{d}{dx}(\tan^{-1}x)\right)\left(\int 1\,dx\right)\,dx\right]$
$\displaystyle =3\left[x\tan^{-1}x-\int \frac{1}{1+x^2}\cdot x\,dx\right]$
$\displaystyle =3x\tan^{-1}x-3\int \frac{x}{1+x^2}\,dx$
$\displaystyle \text{Let }1+x^2=t$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle =3x\tan^{-1}x-\frac{3}{2}\int \frac{dt}{t}$
$\displaystyle =3x\tan^{-1}x-\frac{3}{2}\log|t|+C$
$\displaystyle =3x\tan^{-1}x-\frac{3}{2}\log|1+x^2|+C$

$\displaystyle \textbf{Question 38: }~\int x^2\sin^{-1}x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^2\sin^{-1}x\,dx$
$\displaystyle \text{Taking }\sin^{-1}x\text{ as the first function and }x^2\text{ as the second function.}$
$\displaystyle = \sin^{-1}x\int x^2\,dx-\int \left(\frac{d}{dx}(\sin^{-1}x)\right)\left(\int x^2\,dx\right)\,dx$
$\displaystyle = \sin^{-1}x\cdot \frac{x^3}{3}-\int \frac{1}{\sqrt{1-x^2}}\cdot \frac{x^3}{3}\,dx$
$\displaystyle = \frac{x^3}{3}\sin^{-1}x-\frac{1}{3}\int \frac{x^3}{\sqrt{1-x^2}}\,dx$
$\displaystyle \text{Let }t=1-x^2$
$\displaystyle \Rightarrow dt=-2x\,dx$
$\displaystyle \Rightarrow x\,dx=-\frac{dt}{2}$
$\displaystyle \text{and }x^2=1-t$
$\displaystyle \therefore \int \frac{x^3}{\sqrt{1-x^2}}\,dx=\int \frac{x^2\cdot x}{\sqrt{t}}\,dx=\int \frac{(1-t)}{\sqrt{t}}\left(-\frac{dt}{2}\right)$
$\displaystyle =-\frac{1}{2}\int \left(t^{-1/2}-t^{1/2}\right)dt$
$\displaystyle =-\frac{1}{2}\left[2t^{1/2}-\frac{2}{3}t^{3/2}\right]+C$
$\displaystyle =-\sqrt{t}+\frac{1}{3}t^{3/2}+C$
$\displaystyle =-\sqrt{1-x^2}+\frac{1}{3}(1-x^2)^{3/2}+C$
$\displaystyle \therefore \int x^2\sin^{-1}x\,dx=\frac{x^3}{3}\sin^{-1}x-\frac{1}{3}\left[-\sqrt{1-x^2}+\frac{1}{3}(1-x^2)^{3/2}\right]+C$
$\displaystyle =\frac{x^3}{3}\sin^{-1}x+\frac{\sqrt{1-x^2}}{3}-\frac{(1-x^2)^{3/2}}{9}+C$

$\displaystyle \textbf{Question 39: }~\int \frac{\sin^{-1}x}{x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \frac{\sin^{-1}x}{x^2}\,dx$
$\displaystyle \text{Put }x=\sin\theta$
$\displaystyle \Rightarrow \theta=\sin^{-1}x$
$\displaystyle \Rightarrow dx=\cos\theta\,d\theta$
$\displaystyle \therefore I=\int \frac{\theta\cos\theta}{\sin^2\theta}\,d\theta$
$\displaystyle =\int \theta\left(\frac{\cos\theta}{\sin\theta}\right)\left(\frac{1}{\sin\theta}\right)\,d\theta$
$\displaystyle =\int \theta\,\cot\theta\,\mathrm{cosec}\,\theta\,d\theta$
$\displaystyle \text{Taking }\theta\text{ as the first function and }\cot\theta\,\mathrm{cosec}\,\theta\text{ as the second function.}$
$\displaystyle =\theta\int \cot\theta\,\mathrm{cosec}\,\theta\,d\theta-\int \left(\frac{d}{d\theta}(\theta)\right)\left(\int \cot\theta\,\mathrm{cosec}\,\theta\,d\theta\right)\,d\theta$
$\displaystyle =\theta(-\mathrm{cosec}\,\theta)-\int 1\cdot(-\mathrm{cosec}\,\theta)\,d\theta$
$\displaystyle =-\theta\,\mathrm{cosec}\,\theta+\int \mathrm{cosec}\,\theta\,d\theta$
$\displaystyle =-\theta\,\mathrm{cosec}\,\theta+\log\left|\mathrm{cosec}\,\theta-\cot\theta\right|+C$
$\displaystyle =-\frac{\theta}{\sin\theta}+\log\left|\frac{1-\cos\theta}{\sin\theta}\right|+C$
$\displaystyle =-\frac{\theta}{\sin\theta}+\log\left|\frac{1-\sqrt{1-\sin^2\theta}}{\sin\theta}\right|+C$
$\displaystyle =-\frac{\sin^{-1}x}{x}+\log\left|\frac{1-\sqrt{1-x^2}}{x}\right|+C$

$\displaystyle \textbf{Question 40: }~\int \frac{x^2\tan^{-1}x}{1+x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \frac{x^2\tan^{-1}x}{1+x^2}\,dx$
$\displaystyle =\int \left(\frac{x^2+1-1}{x^2+1}\right)\tan^{-1}x\,dx$
$\displaystyle =\int \left(1-\frac{1}{x^2+1}\right)\tan^{-1}x\,dx$
$\displaystyle =\int 1\cdot \tan^{-1}x\,dx-\int \frac{\tan^{-1}x}{x^2+1}\,dx$
$\displaystyle \text{Taking }\tan^{-1}x\text{ as the first function and }1\text{ as the second function.}$
$\displaystyle =\tan^{-1}x\int 1\,dx-\int \left(\frac{d}{dx}(\tan^{-1}x)\right)\left(\int 1\,dx\right)\,dx-\int \frac{\tan^{-1}x}{x^2+1}\,dx$
$\displaystyle =x\tan^{-1}x-\int \frac{x}{1+x^2}\,dx-\int \frac{\tan^{-1}x}{x^2+1}\,dx$
$\displaystyle \text{Put }t=1+x^2\text{ in the first integral and }p=\tan^{-1}x\text{ in the second integral}$
$\displaystyle \Rightarrow dt=2x\,dx\text{ and }dp=\frac{1}{1+x^2}\,dx$
$\displaystyle \therefore I=x\tan^{-1}x-\frac{1}{2}\int \frac{dt}{t}-\int p\,dp$
$\displaystyle =x\tan^{-1}x-\frac{1}{2}\log|t|-\frac{p^2}{2}+C$
$\displaystyle =x\tan^{-1}x-\frac{1}{2}\log|1+x^2|-\frac{(\tan^{-1}x)^2}{2}+C$

$\displaystyle \textbf{Question 41: }~\int \cos^{-1}(4x^3-3x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \cos^{-1}(4x^3-3x)\,dx$
$\displaystyle \text{Let }x=\cos\theta$
$\displaystyle \Rightarrow \theta=\cos^{-1}x$
$\displaystyle \Rightarrow dx=-\sin\theta\,d\theta$
$\displaystyle \therefore \int \cos^{-1}(4x^3-3x)\,dx=\int \cos^{-1}(4\cos^3\theta-3\cos\theta)(-\sin\theta)\,d\theta$
$\displaystyle =\int \cos^{-1}(\cos 3\theta)(-\sin\theta)\,d\theta$
$\displaystyle =-3\int \theta\sin\theta\,d\theta$
$\displaystyle =-3\left[\theta\int \sin\theta\,d\theta-\int \left(\frac{d}{d\theta}(\theta)\right)\left(\int \sin\theta\,d\theta\right)\,d\theta\right]$
$\displaystyle =-3\left[\theta(-\cos\theta)-\int 1(-\cos\theta)\,d\theta\right]$
$\displaystyle =-3\left[-\theta\cos\theta+\int \cos\theta\,d\theta\right]$
$\displaystyle =-3\left[-\theta\cos\theta+\sin\theta\right]+C$
$\displaystyle =3\theta\cos\theta-3\sin\theta+C$
$\displaystyle =3x\cos^{-1}x-3\sqrt{1-x^2}+C$

$\displaystyle \textbf{Question 42: }~\int \cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right)\,dx$
$\displaystyle =2\int 1\cdot \tan^{-1}x\,dx\quad[\because\ \cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right)=2\tan^{-1}x]$
$\displaystyle =2\left[\tan^{-1}x\int 1\,dx-\int \left(\frac{d}{dx}(\tan^{-1}x)\right)\left(\int 1\,dx\right)\,dx\right]$
$\displaystyle =2\left[x\tan^{-1}x-\int \frac{1}{1+x^2}\cdot x\,dx\right]$
$\displaystyle =2x\tan^{-1}x-\int \frac{2x}{1+x^2}\,dx$
$\displaystyle \text{Put }1+x^2=t$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \therefore I=2x\tan^{-1}x-\int \frac{dt}{t}$
$\displaystyle =2x\tan^{-1}x-\log|t|+C$
$\displaystyle =2x\tan^{-1}x-\log|1+x^2|+C$

$\displaystyle \textbf{Question 43: }~\int \tan^{-1}\!\left(\frac{2x}{1-x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \tan^{-1}\!\left(\frac{2x}{1-x^2}\right)\,dx$
$\displaystyle =2\int 1\cdot \tan^{-1}x\,dx\quad[\because\ \tan^{-1}\!\left(\frac{2x}{1-x^2}\right)=2\tan^{-1}x]$
$\displaystyle =2\left[\tan^{-1}x\int 1\,dx-\int \left(\frac{d}{dx}(\tan^{-1}x)\right)\left(\int 1\,dx\right)\,dx\right]$
$\displaystyle =2\left[x\tan^{-1}x-\int \frac{1}{1+x^2}\cdot x\,dx\right]$
$\displaystyle =2x\tan^{-1}x-\int \frac{2x}{1+x^2}\,dx$
$\displaystyle \text{Put }1+x^2=t$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \therefore I=2x\tan^{-1}x-\int \frac{dt}{t}$
$\displaystyle =2x\tan^{-1}x-\log|t|+C$
$\displaystyle =2x\tan^{-1}x-\log|1+x^2|+C$

$\displaystyle \textbf{Question 44: }~\int (x+1)\log x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int (x+1)\log x\,dx$
$\displaystyle \text{Taking }\log x\text{ as the first function and }(x+1)\text{ as the second function.}$
$\displaystyle = \log x\int (x+1)\,dx-\int \left(\frac{d}{dx}(\log x)\right)\left(\int (x+1)\,dx\right)\,dx$
$\displaystyle = \log x\left(\frac{x^2}{2}+x\right)-\int \frac{1}{x}\left(\frac{x^2}{2}+x\right)\,dx$
$\displaystyle = \log x\left(\frac{x^2}{2}+x\right)-\int \left(\frac{x}{2}+1\right)\,dx$
$\displaystyle = \log x\left(\frac{x^2}{2}+x\right)-\left(\frac{x^2}{4}+x\right)+C$

$\displaystyle \textbf{Question 45: }~\int x^2\tan^{-1}x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int x^2\tan^{-1}x\,dx$
$\displaystyle \text{Taking }\tan^{-1}x\text{ as the first function and }x^2\text{ as the second function.}$
$\displaystyle = \tan^{-1}x\int x^2\,dx-\int \left(\frac{d}{dx}(\tan^{-1}x)\right)\left(\int x^2\,dx\right)\,dx$
$\displaystyle = \tan^{-1}x\cdot \frac{x^3}{3}-\int \frac{1}{1+x^2}\cdot \frac{x^3}{3}\,dx$
$\displaystyle = \frac{x^3}{3}\tan^{-1}x-\frac{1}{3}\int \frac{x^3}{1+x^2}\,dx$
$\displaystyle \text{Let }1+x^2=t$
$\displaystyle \Rightarrow dt=2x\,dx$
$\displaystyle \Rightarrow x\,dx=\frac{dt}{2}$
$\displaystyle \therefore \int \frac{x^3}{1+x^2}\,dx=\int \frac{x^2\cdot x}{t}\,dx=\int \frac{t-1}{t}\cdot \frac{dt}{2}$
$\displaystyle =\frac{1}{2}\int \left(1-\frac{1}{t}\right)dt$
$\displaystyle =\frac{1}{2}\left[t-\log|t|\right]+C$
$\displaystyle \therefore I=\frac{x^3}{3}\tan^{-1}x-\frac{1}{3}\cdot\frac{1}{2}\left[t-\log|t|\right]+C$
$\displaystyle =\frac{x^3}{3}\tan^{-1}x-\frac{t}{6}+\frac{1}{6}\log|t|+C$
$\displaystyle =\frac{x^3}{3}\tan^{-1}x-\frac{1+x^2}{6}+\frac{1}{6}\log|1+x^2|+C$
$\displaystyle =\frac{x^3}{3}\tan^{-1}x-\frac{x^2}{6}+\frac{1}{6}\log|1+x^2|+C$

$\displaystyle \textbf{Question 46: }~\int (e^{\log x}+\sin x)\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int (e^{\log x}+\sin x)\cos x\,dx$
$\displaystyle = \int (x+\sin x)\cos x\,dx\quad(\because\ e^{\log x}=x)$
$\displaystyle = \int x\cos x\,dx+\int \sin x\cos x\,dx$
$\displaystyle = \int x\cos x\,dx+\frac{1}{2}\int 2\sin x\cos x\,dx$
$\displaystyle = \int x\cos x\,dx+\frac{1}{2}\int \sin 2x\,dx$
$\displaystyle = x\int \cos x\,dx-\int \left(\frac{d}{dx}(x)\right)\left(\int \cos x\,dx\right)\,dx+\frac{1}{2}\int \sin 2x\,dx$
$\displaystyle = x\sin x-\int \sin x\,dx+\frac{1}{2}\left(-\frac{\cos 2x}{2}\right)+C$
$\displaystyle = x\sin x+\cos x-\frac{\cos 2x}{4}+C$
$\displaystyle = x\sin x+\cos x-\frac{1}{4}(1-2\sin^2 x)+C$
$\displaystyle = x\sin x+\cos x+\frac{\sin^2 x}{2}-\frac{1}{4}+C$
$\displaystyle = x\sin x+\cos x+\frac{\sin^2 x}{2}+C$

$\displaystyle \textbf{Question 47: }~\int \frac{x\tan^{-1}x}{(1+x^2)^{3/2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \frac{x\tan^{-1}x}{(1+x^2)^{3/2}}\,dx$
$\displaystyle \text{Putting }x=\tan\theta$
$\displaystyle \Rightarrow dx=\sec^2\theta\,d\theta$
$\displaystyle \text{and }\theta=\tan^{-1}x$
$\displaystyle \therefore I=\int \frac{(\tan\theta)\cdot\theta\cdot\sec^2\theta}{(1+\tan^2\theta)^{3/2}}\,d\theta$
$\displaystyle =\int \frac{\theta\tan\theta\sec^2\theta}{(\sec^2\theta)^{3/2}}\,d\theta$
$\displaystyle =\int \frac{\theta\tan\theta\sec^2\theta}{\sec^3\theta}\,d\theta$
$\displaystyle =\int \theta\frac{\tan\theta}{\sec\theta}\,d\theta$
$\displaystyle =\int \theta\sin\theta\,d\theta$
$\displaystyle =\theta\int \sin\theta\,d\theta-\int \left(\frac{d}{d\theta}(\theta)\right)\left(\int \sin\theta\,d\theta\right)\,d\theta$
$\displaystyle =\theta(-\cos\theta)-\int 1(-\cos\theta)\,d\theta$
$\displaystyle =-\theta\cos\theta+\int \cos\theta\,d\theta$
$\displaystyle =-\theta\cos\theta+\sin\theta+C$
$\displaystyle =-\frac{\theta}{\sec\theta}+\frac{1}{\mathrm{cosec}\,\theta}+C$
$\displaystyle =-\frac{\theta}{\sqrt{1+\tan^2\theta}}+\frac{1}{\sqrt{1+\cot^2\theta}}+C$
$\displaystyle =-\frac{\tan^{-1}x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}+C$

$\displaystyle \textbf{Question 48: }~\int \tan^{-1}(\sqrt{x})\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }I=\int \tan^{-1}\sqrt{x}\,dx$
$\displaystyle =\int \frac{\sqrt{x}\cdot \tan^{-1}\sqrt{x}}{\sqrt{x}}\,dx$
$\displaystyle \text{Let }\sqrt{x}=t$
$\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}$
$\displaystyle \Rightarrow \frac{dx}{\sqrt{x}}=2\,dt$
$\displaystyle \therefore I=2\int t\tan^{-1}t\,dt$
$\displaystyle \text{Taking }\tan^{-1}t\text{ as the first function and }t\text{ as the second function.}$
$\displaystyle =2\left[\tan^{-1}t\int t\,dt-\int \left(\frac{d}{dt}(\tan^{-1}t)\right)\left(\int t\,dt\right)\,dt\right]$
$\displaystyle =2\left[\tan^{-1}t\cdot \frac{t^2}{2}-\int \frac{1}{1+t^2}\cdot \frac{t^2}{2}\,dt\right]$
$\displaystyle =t^2\tan^{-1}t-\int \frac{t^2}{1+t^2}\,dt$
$\displaystyle =t^2\tan^{-1}t-\int \left(\frac{1+t^2-1}{1+t^2}\right)dt$
$\displaystyle =t^2\tan^{-1}t-\int 1\,dt+\int \frac{1}{1+t^2}\,dt$
$\displaystyle =t^2\tan^{-1}t-t+\tan^{-1}t+C$
$\displaystyle \text{Substituting }t=\sqrt{x}$
$\displaystyle =x\tan^{-1}\sqrt{x}-\sqrt{x}+\tan^{-1}\sqrt{x}+C$
$\displaystyle =(x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+C$

$\displaystyle \textbf{Question 49: }~\int x^3\tan^{-1}x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^3\tan^{-1}x\,dx$
$\displaystyle \text{Taking }\tan^{-1}x\text{ as the first function and }x^3\text{ as the second function.}$
$\displaystyle = \tan^{-1}x\int x^3\,dx-\int \left(\frac{d}{dx}(\tan^{-1}x)\right)\left(\int x^3\,dx\right)\,dx$
$\displaystyle = \tan^{-1}x\cdot \frac{x^4}{4}-\int \frac{1}{1+x^2}\cdot \frac{x^4}{4}\,dx$
$\displaystyle = \frac{x^4}{4}\tan^{-1}x-\frac{1}{4}\int \frac{x^4}{1+x^2}\,dx$
$\displaystyle = \frac{x^4}{4}\tan^{-1}x-\frac{1}{4}\int \frac{x^4+x^2-x^2-1+1}{1+x^2}\,dx$
$\displaystyle = \frac{x^4}{4}\tan^{-1}x-\frac{1}{4}\int \left(x^2-1+\frac{1}{1+x^2}\right)dx$
$\displaystyle = \frac{x^4}{4}\tan^{-1}x-\frac{1}{4}\int (x^2-1)\,dx-\frac{1}{4}\int \frac{1}{1+x^2}\,dx$
$\displaystyle = \frac{x^4}{4}\tan^{-1}x-\frac{1}{4}\left(\frac{x^3}{3}-x\right)-\frac{1}{4}\tan^{-1}x+C$
$\displaystyle = \left(\frac{x^4}{4}-\frac{1}{4}\right)\tan^{-1}x-\frac{1}{12}(x^3-3x)+C$

$\displaystyle \textbf{Question 50: }~\int x\sin x\cos 2x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x\cos 2x \sin x\,dx$
$\displaystyle = \frac{1}{2}\int x\,(2\cos 2x \sin x)\,dx \quad [\because\ 2\cos A\sin B=\sin(A+B)-\sin(A-B)]$
$\displaystyle = \frac{1}{2}\int x(\sin 3x-\sin x)\,dx$
$\displaystyle = \frac{1}{2}\int x\sin 3x\,dx-\frac{1}{2}\int x\sin x\,dx$
$\displaystyle = \frac{1}{2}\left[x\int \sin 3x\,dx-\int\left(\frac{d}{dx}(x)\int \sin 3x\,dx\right)dx\right]-\frac{1}{2}\left[x\int \sin x\,dx-\int\left(\frac{d}{dx}(x)\int \sin x\,dx\right)dx\right]$
$\displaystyle = \frac{1}{2}\left[x\left(-\frac{\cos 3x}{3}\right)-\int\left(1\cdot -\frac{\cos 3x}{3}\right)dx\right]-\frac{1}{2}\left[x(-\cos x)-\int(1\cdot -\cos x)\,dx\right]$
$\displaystyle = \frac{1}{2}\left[x\left(-\frac{\cos 3x}{3}\right)+\frac{1}{3}\int \cos 3x\,dx\right]-\frac{1}{2}\left[x(-\cos x)+\int \cos x\,dx\right]$
$\displaystyle = \frac{1}{2}\left[-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}\right]-\frac{1}{2}\left[-x\cos x+\sin x\right]$
$\displaystyle = -\frac{x\cos 3x}{6}+\frac{\sin 3x}{18}+\frac{x\cos x}{2}-\frac{\sin x}{2}+C$

$\displaystyle \textbf{Question 51: }~\int (\tan^{-1}x^2)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int x\tan^{-1}x^2\,dx$
$\displaystyle \text{Put } x^2=t$
$\displaystyle \Rightarrow 2x\,dx=dt$
$\displaystyle \Rightarrow I=\frac{1}{2}\int t\tan^{-1}t\,dt$
$\displaystyle =\frac{1}{2}\left[t\int \tan^{-1}t\,dt-\int\left(\frac{dt}{dt}\int \tan^{-1}t\,dt\right)dt\right]$
$\displaystyle =\frac{1}{2}\left[t\left(t\tan^{-1}t-\int\frac{t}{1+t^2}\,dt\right)-\int\left(t\tan^{-1}t-\int\frac{t}{1+t^2}\,dt\right)dt\right]$
$\displaystyle =\frac{1}{2}\left[t^2\tan^{-1}t-\int\frac{t^2}{1+t^2}\,dt\right]$
$\displaystyle =\frac{1}{2}\left[t^2\tan^{-1}t-\int\left(1-\frac{1}{1+t^2}\right)dt\right]$
$\displaystyle =\frac{1}{2}\left[t^2\tan^{-1}t-t+\tan^{-1}t\right]+C$
$\displaystyle =\frac{1}{2}t^2\tan^{-1}t-\frac{1}{2}t+\frac{1}{2}\tan^{-1}t+C$
$\displaystyle =\frac{1}{2}x^2\tan^{-1}x^2-\frac{1}{2}x^2+\frac{1}{2}\tan^{-1}x^2+C$

$\displaystyle \textbf{Question 52: }~\int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx$
$\displaystyle \text{Let } \sin^{-1}x=\theta$
$\displaystyle x=\sin\theta$
$\displaystyle dx=\cos\theta\,d\theta$
$\displaystyle \int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx=\int \frac{(\sin\theta)\theta}{\sqrt{1-\sin^2\theta}}\cdot\cos\theta\,d\theta$
$\displaystyle =\int \frac{\theta\sin\theta}{\cos\theta}\cdot\cos\theta\,d\theta$
$\displaystyle =\int \theta\sin\theta\,d\theta$
$\displaystyle =\theta\int \sin\theta\,d\theta-\int\left(\frac{d\theta}{d\theta}\right)\left(\int \sin\theta\,d\theta\right)d\theta$
$\displaystyle =\theta(-\cos\theta)-\int(1)(-\cos\theta)\,d\theta$
$\displaystyle =-\theta\cos\theta+\sin\theta+C$
$\displaystyle =-\theta\sqrt{1-\sin^2\theta}+\sin\theta+C$
$\displaystyle =-\sin^{-1}x\sqrt{1-x^2}+x+C$

$\displaystyle \textbf{Question 53: }~\int \sin^3\!\sqrt{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \sin^3\sqrt{x}\,dx$
$\displaystyle \text{Put } \sqrt{x}=t$
$\displaystyle \Rightarrow x=t^2$
$\displaystyle \Rightarrow dx=2t\,dt$
$\displaystyle \therefore I=2\int t\sin^3 t\,dt$
$\displaystyle =2\int t\left(\frac{3\sin t-\sin 3t}{4}\right)dt$
$\displaystyle =\frac{1}{2}\int t(3\sin t-\sin 3t)\,dt$
$\displaystyle =\frac{1}{2}\left[3\int t\sin t\,dt-\int t\sin 3t\,dt\right]$
$\displaystyle =\frac{1}{2}\left[3\left(t\int \sin t\,dt-\int\left(\frac{dt}{dt}\int \sin t\,dt\right)dt\right)-\left(t\int \sin 3t\,dt-\int\left(\frac{dt}{dt}\int \sin 3t\,dt\right)dt\right)\right]$
$\displaystyle =\frac{1}{2}\left[3(-t\cos t+\sin t)-\left(-\frac{t\cos 3t}{3}+\frac{1}{9}\sin 3t\right)\right]+C$
$\displaystyle =\frac{1}{2}\left[-3t\cos t+3\sin t+\frac{t\cos 3t}{3}-\frac{1}{9}\sin 3t\right]+C$
$\displaystyle =-\frac{3}{2}t\cos t+\frac{3}{2}\sin t+\frac{1}{6}t\cos 3t-\frac{1}{18}\sin 3t+C$
$\displaystyle =-\frac{3}{2}\sqrt{x}\cos\sqrt{x}+\frac{3}{2}\sin\sqrt{x}+\frac{1}{6}\sqrt{x}\cos(3\sqrt{x})-\frac{1}{18}\sin(3\sqrt{x})+C$

$\displaystyle \textbf{Question 54: }~\int x\sin^3 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int x\sin^3 x\,dx$
$\displaystyle \sin(3A)=3\sin A-4\sin^3 A$
$\displaystyle \sin^3 A=\frac{1}{4}\left(3\sin A-\sin 3A\right)$
$\displaystyle \therefore I=\frac{1}{4}\int x\left(3\sin x-\sin 3x\right)dx$
$\displaystyle =\frac{3}{4}\int x\sin x\,dx-\frac{1}{4}\int x\sin 3x\,dx$
$\displaystyle =\frac{3}{4}\left[x(-\cos x)-\int 1(-\cos x)\,dx\right]-\frac{1}{4}\left[x\left(-\frac{\cos 3x}{3}\right)-\int 1\left(-\frac{\cos 3x}{3}\right)dx\right]$
$\displaystyle =\frac{3}{4}\left[-x\cos x+\sin x\right]-\frac{1}{4}\left[-\frac{x\cos 3x}{3}+\frac{1}{3}\int \cos 3x\,dx\right]$
$\displaystyle =\frac{3}{4}\left[-x\cos x+\sin x\right]-\frac{1}{4}\left[-\frac{x\cos 3x}{3}+\frac{1}{9}\sin 3x\right]+C$
$\displaystyle =-\frac{3x\cos x}{4}+\frac{3\sin x}{4}+\frac{x\cos 3x}{12}-\frac{1}{36}\sin 3x+C$

$\displaystyle \textbf{Question 55: }~\int \cos^3\!\sqrt{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \cos^3\sqrt{x}\,dx \qquad (1)$
$\displaystyle \text{Consider, } \sqrt{x}=t \qquad (2)$
$\displaystyle \text{Differentiating both sides we get,}$
$\displaystyle \frac{1}{2\sqrt{x}}\,dx=dt$
$\displaystyle \Rightarrow dx=2\sqrt{x}\,dt$
$\displaystyle \Rightarrow dx=2t\,dt$
$\displaystyle \text{Therefore, (1) becomes,}$
$\displaystyle I=\int \cos^3 t \cdot 2t\,dt$
$\displaystyle =2\int t\cos^3 t\,dt$
$\displaystyle =2\int t\left(\frac{3\cos t+\cos 3t}{4}\right)dt \quad (\text{Since, } \cos 3A=4\cos^3 A-3\cos A)$
$\displaystyle =\frac{3}{2}\int t\cos t\,dt+\frac{1}{2}\int t\cos 3t\,dt$
$\displaystyle =\frac{3}{2}\left[t\int \cos t\,dt-\int\left(\frac{dt}{dt}\int \cos t\,dt\right)dt\right]+\frac{1}{2}\left[t\int \cos 3t\,dt-\int\left(\frac{dt}{dt}\int \cos 3t\,dt\right)dt\right]$
$\displaystyle =\frac{3}{2}\left[t\sin t-\int \sin t\,dt\right]+\frac{1}{2}\left[t\frac{\sin 3t}{3}-\int \frac{\sin 3t}{3}\,dt\right]$
$\displaystyle =\frac{3}{2}\left[t\sin t+\cos t\right]+\frac{1}{2}\left[\frac{t\sin 3t}{3}+\frac{1}{9}\cos 3t\right]+C$
$\displaystyle =\frac{3}{2}t\sin t+\frac{3}{2}\cos t+\frac{1}{6}t\sin 3t+\frac{1}{18}\cos 3t+C$
$\displaystyle =\frac{3}{2}\sqrt{x}\sin\sqrt{x}+\frac{3}{2}\cos\sqrt{x}+\frac{1}{6}\sqrt{x}\sin(3\sqrt{x})+\frac{1}{18}\cos(3\sqrt{x})+C$

$\displaystyle \textbf{Question 56: }~\int x\cos^3 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int x\cos^3 x\,dx$
$\displaystyle \text{As we know,}$
$\displaystyle \cos 3x=4\cos^3 x-3\cos x$
$\displaystyle \Rightarrow \cos^3 x=\frac{1}{4}(\cos 3x+3\cos x)$
$\displaystyle \therefore I=\frac{1}{4}\int x(\cos 3x+3\cos x)\,dx$
$\displaystyle =\frac{1}{4}\int x\cos 3x\,dx+\frac{3}{4}\int x\cos x\,dx$
$\displaystyle =\frac{1}{4}\left[x\int \cos 3x\,dx-\int\left(\frac{d}{dx}(x)\int \cos 3x\,dx\right)dx\right]+\frac{3}{4}\left[x\int \cos x\,dx-\int\left(\frac{d}{dx}(x)\int \cos x\,dx\right)dx\right]$
$\displaystyle =\frac{1}{4}\left[x\frac{\sin 3x}{3}-\int 1\cdot\frac{\sin 3x}{3}\,dx\right]+\frac{3}{4}\left[x\sin x-\int 1\cdot\sin x\,dx\right]$
$\displaystyle =\frac{x\sin 3x}{12}+\frac{\cos 3x}{36}+\frac{3}{4}x\sin x+\frac{3}{4}\cos x+C$

$\displaystyle \textbf{Question 57: }~\int \tan^{-1}\!\sqrt{\frac{1-x}{1+x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx$
$\displaystyle \text{Putting } x=\cos\theta$
$\displaystyle \Rightarrow dx=-\sin\theta\,d\theta$
$\displaystyle \text{and } \theta=\cos^{-1}x$
$\displaystyle \therefore I=\int \tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-\sin\theta)\,d\theta$
$\displaystyle =\int \tan^{-1}\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}(-\sin\theta)\,d\theta$
$\displaystyle =\int \tan^{-1}\left(\tan\frac{\theta}{2}\right)(-\sin\theta)\,d\theta$
$\displaystyle =-\frac{1}{2}\int \theta\sin\theta\,d\theta$
$\displaystyle =-\frac{1}{2}\left[\theta\int \sin\theta\,d\theta-\int\left(\frac{d\theta}{d\theta}\int \sin\theta\,d\theta\right)d\theta\right]$
$\displaystyle =-\frac{1}{2}\left[\theta(-\cos\theta)-\int 1(-\cos\theta)\,d\theta\right]$
$\displaystyle =-\frac{1}{2}\left[-\theta\cos\theta+\sin\theta\right]+C$
$\displaystyle =\frac{1}{2}\left[\theta\cos\theta-\sin\theta\right]+C$
$\displaystyle =\frac{1}{2}\left[\theta\cos\theta-\sqrt{1-\cos^2\theta}\right]+C$
$\displaystyle =\frac{1}{2}\left[x\cos^{-1}x-\sqrt{1-x^2}\right]+C$
$\displaystyle =\frac{x\cos^{-1}x}{2}-\frac{\sqrt{1-x^2}}{2}+C$

$\displaystyle \textbf{Question 58: }~\int \sin^{-1}\!\sqrt{\frac{x}{a+x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \sin^{-1}\sqrt{\frac{x}{a+x}}\,dx$
$\displaystyle \text{Putting } x=a\tan^2\theta$
$\displaystyle \Rightarrow \sqrt{\frac{x}{a}}=\tan\theta$
$\displaystyle \Rightarrow dx=a(2\tan\theta)\sec^2\theta\,d\theta$
$\displaystyle \therefore I=\int \sin^{-1}\sqrt{\frac{a\tan^2\theta}{a+a\tan^2\theta}}\,(2a\tan\theta\sec^2\theta)\,d\theta$
$\displaystyle =\int \sin^{-1}\sqrt{\frac{\tan^2\theta}{\sec^2\theta}}\,(2a\tan\theta\sec^2\theta)\,d\theta$
$\displaystyle =2a\int \sin^{-1}(\sin\theta)\tan\theta\sec^2\theta\,d\theta$
$\displaystyle =2a\int \theta\tan\theta\sec^2\theta\,d\theta$
$\displaystyle =2a\left[\theta\frac{\tan^2\theta}{2}-\int\frac{\tan^2\theta}{2}\,d\theta\right]$
$\displaystyle =a\left[\theta\tan^2\theta-\int(\sec^2\theta-1)\,d\theta\right]$
$\displaystyle =a\left[\theta\tan^2\theta-\tan\theta+\theta\right]+C$
$\displaystyle =a\theta\tan^2\theta-a\tan\theta+a\theta+C$
$\displaystyle =a\left(\frac{x}{a}\right)\tan^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}+a\tan^{-1}\sqrt{\frac{x}{a}}+C$
$\displaystyle =x\tan^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}+a\tan^{-1}\sqrt{\frac{x}{a}}+C$

$\displaystyle \textbf{Question 59: }~\int \frac{x^3\sin^{-1}x^2}{\sqrt{1-x^4}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{x^3\sin^{-1}x^2}{\sqrt{1-x^4}}\,dx$
$\displaystyle \text{Putting } \sin^{-1}x^2=t$
$\displaystyle \Rightarrow x^2=\sin t$
$\displaystyle \Rightarrow 2x\,dx=\cos t\,dt$
$\displaystyle \Rightarrow \frac{x\,dx}{\sqrt{1-x^4}}=\frac{dt}{2}$
$\displaystyle \therefore I=\int x^2\sin^{-1}x^2\cdot\frac{x\,dx}{\sqrt{1-x^4}}$
$\displaystyle =\int (\sin t)\,t\cdot\frac{dt}{2}$
$\displaystyle =\frac{1}{2}\int t\sin t\,dt$
$\displaystyle =\frac{1}{2}\left[t\int \sin t\,dt-\int\left(\frac{d}{dt}(t)\int \sin t\,dt\right)dt\right]$
$\displaystyle =\frac{1}{2}\left[t(-\cos t)-\int 1(-\cos t)\,dt\right]$
$\displaystyle =\frac{1}{2}\left[-t\cos t+\sin t\right]+C$
$\displaystyle =\frac{1}{2}\left[-t\sqrt{1-\sin^2 t}+\sin t\right]+C$
$\displaystyle =\frac{1}{2}\left[-\sin^{-1}x^2\sqrt{1-x^4}+x^2\right]+C$

$\displaystyle \textbf{Question 60: }~\int \frac{x^2\sin^{-1}x}{(1-x^2)^{3/2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{x^2\sin^{-1}x}{(1-x^2)^{3/2}}\,dx$
$\displaystyle \text{Putting } x=\sin\theta$
$\displaystyle \Rightarrow dx=\cos\theta\,d\theta$
$\displaystyle \text{and } \theta=\sin^{-1}x$
$\displaystyle \therefore I=\int \frac{\sin^2\theta\cdot\theta\cdot\cos\theta\,d\theta}{(1-\sin^2\theta)^{3/2}}$
$\displaystyle =\int \frac{\sin^2\theta\cdot\theta\cdot\cos\theta\,d\theta}{(\cos^2\theta)^{3/2}}$
$\displaystyle =\int \frac{\sin^2\theta\cdot\theta\cdot\cos\theta\,d\theta}{\cos^3\theta}$
$\displaystyle =\int \theta\tan^2\theta\,d\theta$
$\displaystyle =\int (\sec^2\theta-1)\theta\,d\theta$
$\displaystyle =\int \theta\sec^2\theta\,d\theta-\int \theta\,d\theta$
$\displaystyle =\theta\int \sec^2\theta\,d\theta-\int\left(\frac{d}{d\theta}(\theta)\int \sec^2\theta\,d\theta\right)d\theta-\int \theta\,d\theta$
$\displaystyle =\theta\tan\theta-\int 1\cdot\tan\theta\,d\theta-\frac{\theta^2}{2}$
$\displaystyle =\theta\tan\theta-\log|\sec\theta|-\frac{\theta^2}{2}+C$
$\displaystyle =\frac{\theta\sin\theta}{\cos\theta}+\log|\cos\theta|-\frac{\theta^2}{2}+C$
$\displaystyle =\frac{\theta\sin\theta}{\sqrt{1-\sin^2\theta}}+\frac{1}{2}\log|1-\sin^2\theta|-\frac{\theta^2}{2}+C$
$\displaystyle =\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{1}{2}\log(1-x^2)-\frac{1}{2}(\sin^{-1}x)^2+C$