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]$
Best Answer
From the step before the last step you had it as:
$$\eqalign{\int\frac{dx}{\cos x\cos2x} &= \int\frac{dx}{\cos x(2(\cos x)^2 - 1))}\cr & = \int\frac{1}{\cos x} - \frac{2\cos x}{(2(\cos x)^2 - 1)}dx\cr & = \int\frac{1}{\cos x} - \frac{2\cos x}{1 - 2(\sin x)^2}dx\ .\cr}$$
From now you can integrate 1/cosx = secx and its antiderivative is well known.
The second term: let u = sinx, then du = cosxdx, and use fration decomposition to continue.