Question

Evaluate:

${\int }_0^{2\pi }x\left({\sin }^{2}x{\cos }^{2}x\right)dx$

Answer 1

${\int }_0^{2\pi }x{\sin }^2\left(x\right){\cos }^2\left(x\right)dx$

apply integration by parts,

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$u=x,v={\sin }^{2}x{\cos }^{2}x$

$={\left[\frac{1}{32}(x(4x-\sin \left(4x\right))-32\cdot \int \frac{1}{8}(x-\frac{1}{4}\sin \left(4x\right))dx)\right]}_0^{2\pi }$

$={\left[\frac{1}{32}(x(4x-\sin \left(4x\right))-32\cdot \frac{1}{8}(\frac{x^2}{2}+\frac{1}{16}\cos \left(4x\right)))\right]}_0^{2\pi }$

upon simplification, we get,

$={\left[\frac{1}{128}(8x^2-4x\sin \left(4x\right)-\cos \left(4x\right))\right]}_0^{2\pi }$

computing the boundaries, we get,

$=\frac{{\pi }^2}{4}$