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

Question

Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\sin2\text{x}\tan^{-1}(\sin\text{x})\text{dx}$

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Answer

$\text{Let I}=\int^{\frac{\pi}{2}}_{0}\sin2\text{x}\tan^{-1}(\sin\text{x})\text{dx}=\int^{\frac{\pi}{2}}_{0}2\sin\text{x}\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx} $
Also, let $\sin\text{x}=\text{t}=\ \Rightarrow\cos\text{x}\ \text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and when $\text{x}=\frac{\pi}{2},=1$
$\Rightarrow \text{I}=2\int^{1}\limits_{0}\text{t}\cdot\tan^{-1}\text{t dt}$
Consider $\int\text{t}.\tan^{-1}\text{dt}=\tan^{-1}\text{t}.\int\text{t}\ \text{dt}-\int\left\{\frac{\text{d}}{\text{dt}}(\tan^{-1}\text{t)}\int\text{t}\ \text{dt}\right\}\text{dt}$
$=\tan^{-1}\text{t}.\frac{\text{t}^{2}}{2}-\int\frac{1}{1+\text{t}^{2}}.\frac{\text{t}^{2}}{2}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}\int\frac{\text{t}^{2}+1-1}{1+\text{t}^{2}}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}\int1\text{dt}+\frac{1}{2}\int\frac{1}{1+\text{t}^{2}}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}.\text{t}+\frac{1}{2}\tan^{-1}\text{t}$
$\Rightarrow\int^{1}\limits_{0}\text{t}.\tan^{-1}\text{t}\ \text{dt}=\bigg[\frac{\text{t}^{2}.\tan^{-1}\text{t}}{2}-\frac{\text{t}}{2}+\frac{1}{2}\tan^{-1}\text{t}\bigg]^{1}_{0}$
$=\frac{1}{2}\bigg[\frac{\pi}{4}-1+\frac{\pi}{4}\bigg]$
$=\frac{1}{2}\bigg[\frac{\pi}{4}-1\bigg]=\frac{\pi}{4}-\frac{1}{2}$
From equation (1), we obtain
$\text{I}=2\bigg[\frac{\pi}{4}-\frac{1}{2}\bigg]=\frac{\pi}{2}-1$

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