Answer is $A$.
$\int_{0}^{\pi } x^{^{2}} \cos x dx$
$= x^2 \sin x ]_0^{\pi} - \int_0^{\pi} 2x \sin x$
$= x^{^{2}} \sin x ]_0^{\pi} + 2x \cos x ]_0^{\pi}- \int_0^{\pi} 2 \cos x dx$
$= x^{^{2}} \sin x ]_0^{\pi} + 2x \cos x ]_0^{\pi}- 2 \sin x ]_0^{\pi}$
$=[\pi ^2 (0) -0] + 2[ \pi (-1)-0] -2[0-0]$
$=-2\pi$
Integral of a multiplied by b equals a multiplied by integral of b
minus
integral of derivative of a multiplied by integral of b