TITLE: Integrate : $\int \frac{\sin x}{\sin4x} \,dx$
QUESTION [8 upvotes]: Integrate the following  : 
$$\int \frac{\sin x}{\sin4x}dx$$
My approach : 
$$=\int \frac{\sin x}{2\sin2x \cos2x}dx$$
$$= \int \frac{\sin x}{4\sin x \cos x \cos2x}dx$$
$$= \int \frac{1}{4\cos x \cos2x}dx $$ [ didn't get the hint here ] then I did the following way.
$$= \int \frac{\sin x}{4\sin x \cos x (\cos2x) }dx$$
Now putting $\cos x =t$ $\Rightarrow -\sin x \, dx = dt $ 
$$= \int \frac{-dt}{4\sqrt{1-t^2} t(2t^2-1)}$$  [Using : $ \cos2x = 2\cos^2x-1]$

REPLY [9 votes]: $$\frac1{\cos x\cos2x}=\frac{\cos x}{\cos^2x(\cos2x)}=\frac{\cos x}{(1-\sin^2x)(1-2\sin^2x)} $$
Setting $\displaystyle\sin x=u$ in $\displaystyle I=\int\frac{dx}{\cos x\cos2x},$
$\displaystyle I=\int\frac{du}{(1-u^2)(1-2u^2)} =\frac12\int\frac{du}{(1-u^2)\left(\dfrac12-u^2\right)} $
$\displaystyle I=\frac22\int\frac{(1-u^2)-\left(\dfrac12-u^2\right)}{(1-u^2)\left(\dfrac12-u^2\right)}du=\int\frac{du}{1-u^2}-\int\frac{du}{\dfrac12-u^2}$
Can you take it home from here?