Question

${\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx=\frac{2}{3}$

Answer 1

We have $sin3x=3sinx-4si{n}^{3}x$, $si{n}^{3}x=(3sinx-sin3x)/4$

${\int }_0^{\pi /2}si{n}^{3}xdx={\int }_0^{\pi /2}\frac{(3sinx-sin3x)}{4}dx$

$\frac{[-3cosx+\frac{cos3x}{3}{]}_0^{\pi /2}}{4}=\frac{2}{3}$