Integration double angle

Question

How should i simplify this before applying integration. Have tried the $1-\cos2x=2\sin^2x$ but am still stuck on solving it

$$\int\left(\dfrac{\cos2x}{1-\cos4x}\right)dx$$

Answer 1

Using the double angle formula $\cos 4x=1-2\sin^22x$, we have $$ \int \frac{\cos 2x}{1-\cos 4x}dx= \int \frac{\cos 2x}{1-(1-2\sin^22x)}dx=\int \frac{\cos 2x}{2\sin^22x}dx=\frac{1}{2}\int \frac{\cos 2x}{\sin^22x}dx$$

As ABC suggested, use the substitution $y=\sin 2x$, so that $dy=(2\cos 2x)dx\Rightarrow\frac{dy}{2}=(\cos 2x)dx$.

Thus your integral becomes

$$\frac{1}{2}\int \frac{\cos 2x}{\sin^22x}dx=\frac{1}{4}\int\frac{dy}{y^2}$$

This is a relatively simpler integral to evaluate.

Can you take it from here?