# 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 2*2 in denominator is missed ? – Cardinal Jul 17 '15 at 15:47
• @cardinal Actually a 4 – Gummy bears Jul 17 '15 at 15:47
• yep , 2*2 i mean – Cardinal Jul 17 '15 at 15:48
• @Cardinal Fixed it. And BTW did you edit your comment or did it read 2*2 from the beginning?.... – Gummy bears Jul 17 '15 at 15:49
• Use this: u=cos(2x) , and 1+cosx = 2 cos^x /2 – Cardinal Jul 17 '15 at 15:50

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.

• You're welcome. – n55 Jul 17 '15 at 16:01

$$\begin{eqnarray*}\int\frac{\sin(x)\,dx}{\sin(4x)}&=&\frac{1}{4}\int\frac{dx}{2\cos^3 x-\cos x}=\frac{1}{4}\int\frac{\cos x\,dx}{2\cos^4 x-\cos^2 x}\\&=&\frac{1}{4}\int\frac{\cos(x)\,dx}{(1-\sin(x))(1+\sin(x))(1-2\sin^2 x)}\end{eqnarray*}$$ so it is enough to integrate: $$\begin{eqnarray*}\int\frac{dt}{(1-t)(1+t)(1-2t^2)}&=&\int\frac{2\,dt}{1-2t^2}-\int\frac{dt}{1-t^2}\\&=&\sqrt{2}\,\text{arctanh}(\sqrt{2} t)-\text{arctanh}(t)+C.\end{eqnarray*}$$

• This integral should go into the list of evaluations WA doesn't seem to know a simple process for. (Adding it to the polylogarithm list) – Leucippus Jul 17 '15 at 17:11
• @Leucippus Where is this list? I would like to see it's contents. – Jack Tiger Lam Sep 21 '15 at 10:47

\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}

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 $t=\tan(\frac x2)$

• How did you really get those partial fractions? That's the complicated part. – Gummy bears Jul 18 '15 at 4:22
• @Gummybears: in the same way as you would with purely algebraic terms, eg as if the denominator was $x(2x^2-1)$ which can be factorised and split in the usual way. If you want me to edit my answer please let me know, but I think you already have some neater solutions from others :) – David Quinn Jul 18 '15 at 8:06