Question

Evaluate the integrals using substitution.
${\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$

Answer 1

Let $I={\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
Put tan $\theta =x\Rightarrow se{c}^2\theta d\theta =\text{dx or}\theta =ta{n}^{-1}x$
For limit when $x=0\Rightarrow \theta =0\text{and when x =1}\Rightarrow \theta =\frac{\pi }{4}(\because \theta =ta{n}^{-1}x)$
$\therefore I={\int }_0^{\frac{\pi }{4}}si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)se{c}^2\theta d\theta ={\int }_0^{\frac{\pi }{4}}si{n}^{-1}\left(sin2\theta \right)se{c}^2\theta d\theta (\because sin2\theta =\frac{2tan\theta }{1+ta{n}^2\theta })=2{\int }_0^{\frac{\pi }{4}}\theta se{c}^2\theta d\theta$
Now, applying rule of integration by parts taking $\theta$ as the first function and $se{c}^2\theta$ as second function, we get

$I=2{[\theta \int se{c}^2\theta d\theta -\int \left\{\left(\frac{d}{d\theta }\theta \right)\right\}\int se{c}^2\theta d\theta ]}_0^{\frac{\pi }{4}}=2[\theta tan\theta -\int 1tan\theta d\theta {]}_0^{\frac{\pi }{4}}(\because \int se{c}^2\theta d\theta =tan\theta )=2[\theta tan\theta +log|cos\theta |{]}_0^{\frac{\pi }{4}}(\because \int tan\theta d\theta =-logcos\theta )=2\{(\frac{\pi }{4}tan\frac{\pi }{4}+log\left|cos\frac{\pi }{4}\right|)-(0+log|cos0|\left)\right\}=\frac{\pi }{2}+2log\frac{1}{\sqrt{2}}-0(\because log1=0)=\frac{\pi }{2}+2log{2}^{-\frac{1}{2}}=\frac{\pi }{2}-2\times \frac{1}{2}log2(\because log{m}^n=nlogm)=\frac{\pi }{2}-log2$