Evaluate the definite integral: _0^ 2 ^2 x d x - Vidyadip

Question

Evaluate the definite integral: $\int_0^{\frac{\pi}{2}} \cos ^2 x d x$

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Answer

$\int_0^{\frac{\pi}{2}} \cos ^2 x d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{2}}(1+\cos 2 x) d x$
$\because \cos 2 x=1-2 \cos ^2 x$
$\Rightarrow \cos ^2 x=\frac{1}{2}(1+\cos 2 x)$
$=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}$
$=\frac{1}{2}\left[\frac{\pi}{2}-\frac{\sin \pi}{2}-0-\frac{1}{2} \sin 0\right]$
$=\frac{1}{2}\left[\frac{\pi}{2}+0-0\right]$
$=\frac{\pi}{4}$

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