Evaluate: _0^ 2 ^2 x (1+ x)^2 d x

Question

Evaluate: $\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{(1+\cos x)^2} d x$

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Answer

Let $I =\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{(1+\cos x)^2} d x$
Put $\tan \left(\frac{x}{2}\right)= t$
$\therefore x =2 \tan ^{-1} t$
$\therefore dx =\frac{2 dt }{1+ t ^2}, \sin x \frac{2 t }{1+ t ^2} \text { and } x =\frac{1- t ^2}{1+ t ^2}$
When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$
$ \therefore I=\int_0^1 \frac{\left(\frac{2 t }{1+ t ^2}\right)^2}{\left(1+\frac{1- t ^2}{1+ t ^2}\right)^2} \cdot \frac{2 dt }{1+ t ^2}$
$=\int_0^1 \frac{\frac{4 t ^2}{\left(1+ t ^2\right)^2}}{\frac{4}{\left(1+ t ^2\right)^2}} \cdot \frac{2 dt }{1+ t ^2}$
$=2 \int_0^1 \frac{ t ^2}{1+ t ^2} dt$
$=2 \int_0^1\left(\frac{1+ t ^2-1}{1+ t ^2}\right) dt$
$=2 \int_0^1\left(1+\frac{1}{1+ t ^2}\right) dt$
$=2\left[ t -\tan ^{-1} t \right]_0^1$
$=2\left[\left(1-\tan ^{-1} 1\right)-\left(0-\tan ^{-1} 0\right)\right]$
$=2\left(1-\frac{\pi}{4}\right)$
$=\frac{4-\pi}{2} $

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