Question

Evaluate : $\int \frac{sinx}{sin4x}dx$

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

Answer 1

Answer: (A)

We have,

$I=\int \frac{sinx}{sin4x}dx=\int \frac{sinx}{2sin2xcos2x}dx$

$=\int \frac{sinx}{4sinxcosxcos2x}dx$

$\Rightarrow I=\frac{1}{4}\int \frac{1}{cosxcos2x}dx=\frac{1}{4}\int \frac{cosx}{co{s}^{2}xcos2x}dx$

$\Rightarrow I=\frac{1}{4}\int \frac{cosx}{\left(1-si{n}^{2}x\right)\left(1-2si{n}^{2}x\right)}dx$

putting$sinx=t\Rightarrow cosxdx=dt$, 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 $y=1andy=\frac{1}{2}respectivelyin\left(i\right),wegetA=-1andB=2$

$\therefore \frac{1}{\left(1-y\right)\left(1-2y\right)}=-\frac{1}{1-y}+\frac{2}{1-2y}$

$\Rightarrow \frac{1}{\left(1-t^2\right)\left(1-2t^2\right)}=-\frac{1}{1-t^2}+\frac{2}{1-2t^2}$

$\Rightarrow I=\frac{1}{4}\int \left(-\frac{1}{1-t^2}+\frac{2}{1-2t^2}\right)dt$

$\Rightarrow I=-\frac{1}{4}\cdot \frac{1}{2}log\left|\frac{1+t}{1-t}\right|+\frac{1}{2}\cdot \frac{1}{2\sqrt{2}}log\left|\frac{1+\sqrt{2t}}{1-\sqrt{2t}}\right|+C$

$\Rightarrow I=-\frac{1}{8}\left|\frac{1+sinx}{1-sinx}\right|+\frac{1}{4\sqrt{2}}log\left|\frac{1+\sqrt{2sinx}}{1-\sqrt{2sinx}}\right|+C$