Question

$\int x^2\sin 2xdx$$=-\frac{1}{k}{x}^m\cos 2x+\frac{1}{n}x\sin 2x+\frac{1}{p}\cos 2x.$.Find$k+m+n+p$ ?

Answer 1

$\int x^2\sin 2xdx=x^2(-\frac{\cos 2x}{2})-\int 2x(-\frac{\cos 2x}{2})dx$
$=x^2(-\frac{\cos 2x}{2})-2x(-\frac{\sin 2x}{4})+\int 2(-\frac{\sin 2x}{4})dx$
$=x^2(-\frac{\cos 2x}{2})-2x(-\frac{\sin 2x}{4})+2\left(\frac{\cos 2x}{8}\right)$
$=\frac{1}{2}{x}^2\cos 2x+\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x\therefore k+m+n+p=10$