Evaluate: _0^ ^2 x d x - Vidyadip

Question

Evaluate: $\int_0^\pi \cos ^2 x d x$

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Answer

$\int_0^\pi \cos ^2 x d x=\int_0^\pi\left(\frac{1+\cos 2 x}{2}\right) d x$
$=\frac{1}{2}\left[\int_0^\pi d x+\int_0^\pi \cos 2 x d x\right]$
$=\frac{1}{2}\left[[x]_0^\pi+\left[\frac{\sin 2 x}{2}\right]_0^\pi\right]$
$=\frac{1}{2}\left[(\pi-0)+\frac{1}{2}(\sin 2 \pi-\sin 0)\right]$
$=\frac{1}{2}\left[\pi+\frac{1}{2}(0-0)\right]$
$=\frac{\pi}{2}$

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