Question

Evaluate : $\int\frac{sin x}{sin 4x} dx$

• A. $-\frac{1}{8}log\left|\frac{1+sin x}{1-sin x}\right|+\frac{1}{4\sqrt{2}}log \left|\frac{1+\sqrt{2}sin x}{1-\sqrt{2}sin x}\right| +C $
• B. $\frac{1}{8}log\left|\frac{1+sin x}{1-sin x}\right|-\frac{1}{4\sqrt{2}}log \left|\frac{1+\sqrt{2}sin x}{1-\sqrt{2}sin x}\right| +C $
• C. $-\frac{1}{8}log\left|\frac{1+sin x}{1-sin x}\right|+\frac{1}{4\sqrt{2}}log \left|\frac{1-\sqrt{2}sin x}{1+\sqrt{2}sin x}\right| +C $
• D. None of these

Answer 1

The Correct Option is A

We have, $ I= \int\frac{sinx}{sin 4x} dx = \int\frac{sinx}{2sin 2x cos2x} dx$ $ =\int\frac{sinx}{4sinx cosx cos2x }dx$ $\Rightarrow I= \frac{1}{4}\int\frac{1}{cosx cos2x }dx =\frac{1}{4} \int\frac{cosx}{cos^{2}x cos 2x}dx$ $\Rightarrow I =\frac{1}{4} \int\frac{cosx}{\left(1-sin^{2}x\right)\left(1-2sin^{2}x\right)}dx$ putting$ sinx = t \Rightarrow cosx dx = dt$, we get $ I= \frac{1}{4}\int\frac{dt}{\left(1-t^{2}\right)\left(1-2t^{2}\right)}$ $Let t^{2}=y$. Then, $ \frac{1}{\left(1-t^{2}\right)\left(1-2t^{2}\right)}=\frac{1}{\left(1-y\right)\left(1-2y\right)}$ and $\frac{1}{\left(1-y\right)\left(1-2y\right)}= \frac{A}{1-y} +\frac{B}{1-2y}$. $ \Rightarrow 1 =A\left(1-2y\right)+B\left(1-y\right) .......\left(i\right) $ Putting $y=1 and y=\frac{1}{2} respectively in \left(i\right), we get A = -1 and B = 2$ $\therefore\frac{1}{\left(1-y\right)\left(1-2y\right)}= -\frac{1}{1-y} +\frac{2}{1-2y} $ $\Rightarrow \frac{1}{\left(1-t^{2}\right)\left(1-2t^{2}\right)}=-\frac{1}{1-t^{2}}+\frac{2}{1-2t^{2}}$ $ \Rightarrow I= \frac{1}{4}\int \left(-\frac{1}{1-t^{2}}+\frac{2}{1-2t^{2}}\right)dt $ $\Rightarrow I =-\frac{1}{4}\cdot\frac{1}{2}log\left|\frac{1+t}{1-t}\right| +\frac{1}{2}\cdot\frac{1}{2\sqrt{2}}log\left|\frac{1+\sqrt{2t}}{1-\sqrt{2t}}\right|+ C $ $\Rightarrow I = -\frac{1}{8}\left|\frac{1+sinx}{1-sinx }\right| +\frac{1}{4\sqrt{2}}log\left|\frac{1+\sqrt{2 sinx}}{1-\sqrt{2 sinx}}\right| +C $