Question

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

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx$
$I={\int }_0^{\frac{\pi }{2}}{\sin }^{2}x\cdot \sin xdx$
$={\int }_0^{\frac{\pi }{2}}(1-{\cos }^{2}x)\sin xdx$
$={\int }_0^{\frac{\pi }{2}}\sin xdx-{\int }_0^{\frac{\pi }{2}}{\cos }^{2}x\cdot \sin xdx$
Put $\cos x=t$ in second part $\Rightarrow -\sin xdx=dt$
$\therefore I=[-\cos x{]}_0^{\frac{\pi }{2}}-{\left[\frac{t^3}{3}\right]}_0^1$
$=1-\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3}$
Hence, the given result is proved.