Integration of $\frac{\sin x}{\sin 4x}$ 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?
 A: 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.
A: $$\begin{eqnarray*}\int\frac{\sin(x)\,dx}{\sin(4x)}&=&\frac{1}{4}\int\frac{dx}{2\cos^3 x-\cos x}=\frac{1}{4}\int\frac{\cos x\,dx}{2\cos^4 x-\cos^2 x}\\&=&\frac{1}{4}\int\frac{\cos(x)\,dx}{(1-\sin(x))(1+\sin(x))(1-2\sin^2 x)}\end{eqnarray*}$$
so it is enough to integrate:
$$\begin{eqnarray*}\int\frac{dt}{(1-t)(1+t)(1-2t^2)}&=&\int\frac{2\,dt}{1-2t^2}-\int\frac{dt}{1-t^2}\\&=&\sqrt{2}\,\text{arctanh}(\sqrt{2} t)-\text{arctanh}(t)+C.\end{eqnarray*}$$
A: $$\eqalign{\frac{1}{4}\int\frac{dx}{\cos x\cos2x} &= \frac{1}{4}\int\frac{dx}{\cos x(2(\cos x)^2 - 1))}\cr
  & =
\frac{1}{4}\int\frac{1}{\cos x} - \frac{2\cos x}{(2(\cos x)^2 - 1)}dx\cr
  & = \frac{1}{4}\int\frac{1}{\cos x} - \frac{2\cos x}{1 - 2(\sin x)^2}dx\ .\cr}$$
A: You can split the integrand into partial fractions as $$\frac 14\left(\frac{-1}{\cos x}+\frac{1}{2\cos x-\sqrt2}+\frac{1}{2\cos x+\sqrt2}\right)$$
Then use the usual substitution $t=\tan(\frac x2)$
