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

Question

Evaluate the integral in Exercise:

$\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}}{1+\cos^{2}\text{x}}\text{dx}$

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Answer

$\text{Let}\ \text{I}=\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}}{1+\cos^{2}\text{x}}\text{dx}$
$\text{put}\ \cos\text{x}=\text{t},\therefore\sin\text{x}\ \text{dx}=\text{dt}\Rightarrow\sin\text{x}\ \text{dx}=-\text{dt}\ \text{when}\ \text{x}=0,\text{t}=\cos0=1$
$\text{when}\ \text{x}=\frac{\pi}{2},\text{t}=\cos\frac{\pi}{2}=0$
$\therefore\ \ \text{I}=-\int\limits^{0}_{1}\frac{\text{dt}}{1+\text{t}^{2}}=-\big[\tan^{-1}\text{t}\big]^{0}_{1}=-\big[\tan^{-1}0-\tan^{-1}1\big]=-\bigg[0-\frac{\pi}{4}\bigg]=\frac{\pi}{4}$

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