# Evaluating $\int \frac{\sin x}{\sin 4x}\mathrm dx$ [duplicate]

$$\int \frac{\sin x}{\sin 4x}\mathrm dx$$ I have tried to do this by expanding $\sin 4x$ but it was worthless.

• Sorry Adesh Tamrakar and John.. I did not noticed that.. and i have typed solution. – juantheron Sep 21 '15 at 19:43
• @juantheron, no worries. – John_dydx Sep 21 '15 at 19:50

Using $\displaystyle \bullet\; \sin 2x = 2\sin x\cdot \cos x$ and $\; \bullet\; \cos 2x = 1-2\sin^2 x$

Let $$\displaystyle I = \int\frac{\sin x}{2\sin 2x\cdot \cos 2x}dx = \int\frac{\sin x}{4\sin x\cos x\cdot \cos 2x}dx = \frac{1}{4}\int\frac{1}{\cos x\cdot \cos 2x}dx$$

Now $$\displaystyle I = \frac{1}{4}\int\frac{\cos x}{\cos^2 x\cdot (1-2\sin^2x)}dx = \frac{1}{4}\int\frac{\cos x}{(1-\sin^2x)\cdot (1-2\sin^2x)}dx$$

Now Put $\sin x= t\;,$ Then $\cos xdx = dt$

So we get $$\displaystyle I = \frac{1}{4}\int\frac{1}{(1-t^2)(1-2t^2)}dt = \frac{1}{4}\int\frac{1}{(2t^2-1)(t^2-1)}dt$$

So we get $$\displaystyle I = -\frac{1}{4}\int\left[\frac{2}{(2t^2-1)}-\frac{1}{t^2-1}\right] = -\frac{1}{4}\int\frac{1}{t^2-\left(\frac{1}{\sqrt{2}}\right)^2}dt+\frac{1}{4}\int\frac{1}{t^2-1}dt$$

So we get $$\displaystyle I = -\frac{1}{4}\cdot \frac{1}{\sqrt{2}}\ln\left|\frac{\sqrt{2}t-1}{\sqrt{2}t+1}\right|+\frac{1}{4}\cdot \frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+\mathcal{C}$$

So we get $$\displaystyle I = -\frac{1}{4\sqrt{2}}\ln\left|\frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1}\right|+\frac{1}{8}\ln\left|\frac{\sin x-1}{\sin x+1}\right|+\mathcal{C}$$