Question:
Solve the following integral: $$\int \frac{\sin x}{\sin4x}dx$$
Attempt:
Using trigonometric identities to expand $\sin4x$, I obtained the integral:
$$\int \frac{1}{4\cos x \cos2x}dx$$
Now I'm not sure how to proceed. I tried writing $\cos2x$ in terms of $\sin x$ and $\cos x$ however that wasn't helpful. Is there a certain substitution to make? Also, is it possible to use partial fractions when dealing with trigonometric functions?
Best Answer
Substitute $\cos 2x=1-2\sin^2(x)$ and multiply by $\cos(x)$ on top and bottom, and then let $u=\sin(x)$ and $du=\cos(x)dx$: $$\frac{1}{4}\int \frac{1}{\cos x \cos 2x} dx\\ =\frac{1}{4}\int \frac{\cos x}{\cos^2 x (1-2\sin^2(x))} dx\\ =\frac{1}{4}\int \frac{\cos x}{(1-\sin^2 x) (1-2\sin^2(x))} dx\\ =\frac{1}{4}\int \frac{1}{(1-u^2) (1-2u^2)} du\\ $$
Then use partial fractions.