Question

Solve:
${\int }_0^{\frac{\pi }{2}}\left(\sin 2x\right)\sin xdx$

Answer 1

$I={\int }_0^{\pi /2}\left(\sin 2x\right)\sin xdx$

$={\int }_0^{\pi /2}\left(2\sin x\cos x\right)\sin xdx$

$={\int }_0^{\pi /2}2{\sin }^{2}x\cos xdx$

Let $\sin x=t$

$\cos xdx=dt$

$I={\int }_0^{1}2t^{2}dt$

$=2\times \frac{t^3}{3}{|}_0^1$

$=\frac{2}{3}(1^3-0^3)=\frac{2}{3}$.