Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 531: 57

$$\int_{-\pi}^{\pi} \sin ^{2} x d x =\pi $$

$$ \int_{-\pi}^{\pi} \sin ^{2} x d x $$ Since $$\sin^2x= \frac{1}{2}(1-\cos 2x)$$ Then \begin{align*} \int_{-\pi}^{\pi} \sin ^{2} x d x&=\frac{1}{2}\int_{-\pi}^{\pi} (1-\cos2 x) d x\\ &=\frac{1}{2}\left(x-\frac{1}{2}\sin 2x\right)\bigg|_{-\pi}^{\pi}\\ &=\frac{1}{2} (\pi -(-\pi ))\\ &=\pi \end{align*}