Question

\[\int\limits_{- \pi/2}^{\pi/2} \sin^2 x\ dx .\]

Answer 1

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^2 x\ d x\]

\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]

\[ = \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 1 - \cos2x \right)dx\]

\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]

\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 + \frac{\pi}{2} - 0 \right)\]

\[ = \frac{\pi}{2}\]