Question

Evaluate :
${\int }_0^{\pi /2}{x}^{2}sinxdx$

Answer 1

${\int }_0^{\pi /2}{x}^{2}sinx$ dx
Integrating by parts we get,
$=x^2\int \sin xdx-\int [\frac{d}{dx}{x}^2\int \sin xdx]dx$
$={-x}^2\cos x+2\int x\cos xdx$
$=-x^2\cos x+2[x\int \cos xdx-\int [\frac{d}{dx}x\int \cos xdx]dx]$
$=-x^2\cos x+2[x\sin x-\int \sin xdx]$
$={[-x^2\cos x+2x\sin x+2\cos x]}_0^{\pi /2}$
$=\pi -2$