Answer
$\frac{1}{2}$
Work Step by Step
Recall the identity $2\sin x\cos x=\sin2x$
Therefore, we get
$\int \sin x \cos xdx= \frac{1}{2}\int \sin 2xdx$
$=\frac{1}{2}\times\frac{-\cos 2x}{2}+C= -\frac{1}{4}\cos 2x+C$
$\int^{\pi/2}_{0}\sin x\cos xdx= [-\frac{1}{4}\cos 2x]^{\pi/2}_{0}$
$= -\frac{1}{4}\cos\pi-(-\frac{1}{4}\cos 0)$$=(-\frac{1}{4}\times-1)-(-\frac{1}{4}\times1)$
$= \frac{1}{2}$
