Integration of $\int \frac{\cos2x}{\cos^2x} \ dx$

Question

From the integral $$\int \frac{\cos2x}{\cos^2x} \ dx$$ I managed to get $$\int \frac{2\cos2x}{\cos2x+1}\ dx $$ How do I simplify it further?

Answer 1

HINT:

According to André Nicolas we have \begin{align} \int\frac{\cos 2x}{\cos^2 x}\,dx&=\int\frac{2\cos^2 x-1}{\cos^2 x}\,dx=\int(2-\sec^2 x)\,dx \end{align}

Answer 2

Hint: $$\cos 2x=\cos^2x-\sin^2x$$