Question

Integrate $\int x.sin2xdx$

Answer 1

$\int x.\sin 2xdx=?$

Using ILATE function, we have

$\int \underset{I}{x}.\underset{II}{\sin 2x}dx$

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

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

$=-\frac{x\cos 2x}{2}+\frac{1}{2}\left(\frac{\sin 2x}{2}\right)+C$

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

Hence $\int x.\sin 2xdx=\frac{\sin 2x}{4}-\frac{x\cos 2x}{2}+C$.