$\displaystyle\int_0^{\pi/2} \sin^2 x\, dx = ?$
- $\pi/4$
- $\pi/2$
- $1$
- $\pi^2/8$
Show answer and solution
Correct answer: A. $\pi/4$
$\sin^2 x = (1 - \cos 2x)/2$.
$\int_0^{\pi/2} (1 - \cos 2x)/2\, dx = \tfrac{1}{2}[x - \tfrac{1}{2}\sin 2x]_0^{\pi/2} = \tfrac{1}{2}(\pi/2 - 0) = \pi/4$.