Evaluate: _ 6 ^ 3 ^2 x d x

Question

Evaluate: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin ^2 x d x$

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Answer

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin ^2 x d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{1-\cos 2 x}{2}\right) d x$
$=\frac{1}{2}\left[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x-\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos 2 x d x\right]$
$=\frac{1}{2}\left[[x]_{\frac{\pi}{3}}^{\frac{\pi}{6}}-\left[\frac{\sin 2 x}{2}\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\right]$
$=\frac{1}{2}\left[\left(\frac{\pi}{3}-\frac{\pi}{6}\right)-\frac{1}{2}\left(\sin \frac{2 \pi}{3}-\sin \frac{\pi}{3}\right)\right]$
$=\frac{1}{2}\left[\frac{\pi}{6}-\frac{1}{2}\left(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\right)\right]$
$=\frac{1}{2}\left[\frac{\pi}{6}-\frac{1}{2}(0)\right]$
$=\frac{\pi}{12}$

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