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

Answer: (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 $