Evaluate the integral in Exercise: ^ 1 _ 0 ^ -1 ( 2x 1+x^ 2 )dx

Question

Evaluate the integral in Exercise:
$\int\limits^{1}_{0}\sin^{-1}\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)\text{dx}$

✓

Answer

$\text{Let}\text{I}=\int\limits_{0}^{1}\sin^{-1}\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)\text{dx}$
$\text{put}\ \text{x}=\tan\theta\ \text{so that}\ \text{dx}=\sec^{2}\theta\ \text{d}\theta\ \text{when}\ \text{x}=0,\tan\theta=0\Rightarrow\theta=0$
$\text{when}\ \text{x}=1,\tan\theta=1\ \Rightarrow\theta=\frac{\pi}{4}$
$\therefore |=\int^{\frac{\pi}{4}}_{0}\sin^{-1}\bigg(\frac{2\tan\theta}{1+\tan^{2}\theta}\bigg).\sec^{2}\theta\ \text{d}\theta$
$ =\int^{\frac{\pi}{4}}_{0}\sin^{-1}(\sin2\theta).\sec^{2}\theta\ \text{d}\theta=\int^{\frac{\pi}{4}}_{0}2\theta.\sec^{2}\theta\ \text{d}\theta=2\int^{\frac{\pi}{4}}_{0}\theta.\sec^{2}\theta\ \text{d}\theta$
$=2\left\{[\theta\tan\theta]^{\frac{\pi}{4}}_{1}-\int^{\frac{\pi}{4}}_{0}1.\tan\theta\ \text{d}\theta\right\}=2\left\{[\theta\tan\theta]^{\frac{\pi}{4}}_{0}+[\log\cos\theta]^{\frac{\pi}{4}}_{0}\right\}$
$=2\left\{\frac{\pi}{4}\tan\frac{\pi}{4}-0+\log\cos\frac{\pi}{4}-\log\cos0\right\}=2\bigg[\frac{\pi}{4}\times1-0+\log\frac{1}{\sqrt{2}}-\log1\bigg]$
$=2\bigg[\frac{\pi}{4}+\log1-\log\sqrt{2}\bigg]=\frac{\pi}{2}-\log2$

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