How do I integrate $\frac{\sin x}{ \cos x+\cos 2x} \,dx$

Question

I need to evaluate : $$ I=\int \dfrac {\sin x }{ \cos x+\cos 2x}\, dx$$

I tried to write $\cos 2x$ as $\cos^2x-\sin^2x$ then use $\sin^2x+\cos^2x=1$ but I couldn't do any substitution, how can I integrate this.

Answer 1

Hint:

$$\int\frac{\sin x\mathrm dx}{\cos x+\cos 2x}=\int\frac{\sin x\mathrm dx}{2\cos^2x+\cos x-1}=\int\frac{\sin x\mathrm dx}{(2\cos x-1)(\cos x+1)}$$

Set $u=\cos x$, $ \mathrm du =-\sin x\mathrm dx$. Can you proceed?

Answer 2

Hint:

Bioche's rules suggest the substitution $u=\cos x$, $\mathrm du=-\sin x\,\mathrm dx$ . You obtain, using the duplication formula for $\cos 2x$, $$ I=\int \frac {\sin x }{\cos x+\cos 2x}\,\mathrm dx=\int\frac{-\mathrm d u}{u+2u^2-1}=\int\frac{\mathrm d u}{(1+u)(1-2u)}.$$ There remains to decompose this fraction into simple fractions, integrate and revert to the initial variable.