Question

Integrate with respect to $x$:
$x{\sin }^{2}x$

Answer 1

We have,

$\int x{\sin }^{2}xdx$

Using formula,

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

Then,

$\int x{\sin }^{2}xdx=\int x\frac{(1-\cos 2x)}{2}dx$

$=\int \frac{x}{2}dx-\frac{1}{2}\int x.\cos 2xdx$

$=\frac{x^2}{4}-\frac{1}{2}[x\int \cos 2xdx-\int \left(\frac{dx}{dx}\cos 2xdx\right)dx]$

$=\frac{x^2}{4}-\frac{1}{2}[x\frac{\sin 2x}{2}-\frac{\sin 2x}{2}]+C$

$=\frac{x^2}{4}-\frac{1}{2}[x\frac{\sin 2x}{2}-\frac{1}{2}\int \sin 2xdx]+C$

$=\frac{x^2}{4}-\frac{1}{2}[x\frac{\sin 2x}{2}+\frac{1}{4}\cos 2x]+C$

$=\frac{x^2}{4}-\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+C$

Hence, this is the answer.