Answer
$\pi^2 -4$
Work Step by Step
Our aim is to solve the integral $ \int_0^{\pi} x^{2} \sin x dx$
Use integration by parts formula.
$ \int x^{2} \sin x dx =-x^2 \cos x +\int 2x \cos x \ dx \\= -x^2 \cos x +2x \sin x -2\int \sin x +C\\ = -x^2 \cos x +2x \sin x +2\cos x +C$
Now, $ \int_0^{\pi} x^{2} \sin x dx=[-x^2 \cos x +2x \sin x +2\cos x +C]_0^{\pi}\\=-\pi^2 (-1) +2 \times (-1) -2 \\=\pi^2 -4$
