Integrals — Class 12 Maths Solution

: Let $\int\limits_0^{\pi /2} {{{\sin }^3}xdx} = \cfrac{1}{4}\int\limits_0^{\pi /2} {\left( {3\sin x - \sin 3x} \right)dx}$

$= \cfrac{1}{4}\left[ { - 3\cos x + \cfrac{{\cos 3x}}{3}} \right]_0^{\pi /2}$

$= \cfrac{1}{4}\left[ { - 3\cos \cfrac{\pi }{2} + \cfrac{1}{3}\cos \cfrac{{3\pi }}{2}} \right] - \cfrac{1}{4}\left[ { - 3\cos 0 + \cfrac{{\cos 0}}{3}} \right]$

$= \cfrac{1}{4}\left[ {0 + 0 + 3 - \cfrac{1}{3}} \right] = \cfrac{1}{4}\left( {\cfrac{8}{3}} \right) = \cfrac{2}{3}$