solving $\int \cos^2x \sin2x dx$ 
$$\int \cos^2x \sin2x dx$$

$$\int \cos^2x \sin2x \, dx=\int \left(\frac{1}{2} +\frac{\cos2x}{2} \right) \sin2x \, dx$$
$u=\sin2x$
$du=2\cos2x\,dx$
$$\int \left(\frac{1}{2} +\frac{du}{4}\right)u \, du$$
Is the last step is ok?

General strategy in such cases (trigonometric functions, multiples of some angles) is to express everything in terms of $\sin x$ and $\cos x$, and take it from there.
