$\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \mathrm{~d} x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(1-\cos 2 x) d x=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}ight]_{0}^{\frac{\pi}{2}} \ &=\frac{1}{2}\left[\frac{\pi}{2}+0ight]=\frac{\pi}{4} \end{aligned}$

$\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \mathrm{~d} x=$

Solution