Question

Solve:

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^2\sin xdx$

Answer 1

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^2\sin xdx$

$x^2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin xdx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}2x\cos xdx$

$x^2{{\left[-\cos x\right]}_0}^{\frac{\pi }{2}}+{{\left(2x\sin x\right)}_0}^{\frac{\pi }{2}}-\underset{0}{\overset{\frac{\pi }{2}}{\int }}2\sin xdx$

$-\frac{\pi }{4}\left[\cos \frac{\pi }{2}-\cos 0\right]+\left(2\frac{\pi }{2}\sin \frac{\pi }{2}-0\right)+{\left(2\cos x\right)}^{\frac{\pi }{2}}$

$-\frac{\pi }{4}\left[0-1\right]+\pi +2\left(\cos \frac{\pi }{2}-\cos 0\right)$

$\frac{{\pi }^2}{4}+\pi +2\left(0-1\right)$

$\frac{{\pi }^2}{4}+\pi -2$