Question

The value of $\int \frac{\sin x}{\sin 4x}dx$ is

• A. $\frac{1}{4}\log |\frac{{\sin }_X-1}{{\sin }_X+1}|-\frac{1}{\sqrt{2}}\log |\frac{\sqrt{2}{\sin }_X-1}{\sqrt{2}{\sin }_X+1}|+c$
• B. $\frac{1}{8}\log |\frac{\cos x-1}{\cos x+1}|-\frac{1}{2\sqrt{2}}\log |\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}|+c$
• C. $\frac{1}{8}\log |\frac{\sin x-1}{\sin x+1}|-\frac{1}{4\sqrt{2}}\log |\frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1}|+c$
• D. $\frac{1}{8}\log |\frac{\sin x-1}{\sin x+1}|+\log |\frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1}|+c$

Answer 1

The correct option is C $\frac{1}{8}\log |\frac{\sin x-1}{\sin x+1}|-\frac{1}{4\sqrt{2}}\log |\frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1}|+c$

$I=\int \frac{\sin x}{\sin 4x}dx$
$=\frac{1}{4}\int \frac{1}{\cos x({\cos }^{2}x-{\sin }^{2}x)}dx$
$=\frac{1}{4}\int \frac{\cos x}{(1-{\sin }^{2}x)(1-2{\sin }^{2}x)}dx$
Put $\sin x=t\Rightarrow \cos xdx=dt$
$\Rightarrow I=\frac{1}{4}\int \frac{dt}{(1-t^2)(1+t^2)}$
Now using partial fraction method,we will find the value of
$\frac{1}{(1-t)(1+t)(1-t^2)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1-\sqrt{2}t}+\frac{D}{1+\sqrt{2}t}$
$\Rightarrow 1=A(1+t)(1-2t^2)+B(1-t)(1-2t^2)+C(1-t^2)(1+\sqrt{2}t)+D(1-t^2)(1-\sqrt{2}t)$
For $t=1\Rightarrow A=-\frac{1}{2}$
For $t=-1\Rightarrow B=-\frac{1}{2}$
For $t=0\Rightarrow A+B+C+D=1$
$\Rightarrow C+D=2$
For $t=\frac{1}{\sqrt{2}}\Rightarrow C=1$
$\Rightarrow D=1$
So,$I=\frac{1}{4}[-\frac{1}{2}\int \frac{dt}{1-t}-\frac{1}{2}\int \frac{dt}{1+t}+\int \frac{dt}{1-\sqrt{2}t}+\int \frac{dt}{1+\sqrt{2}t}]$
$=\frac{1}{8}log|1-\sin x|-\frac{1}{8}log|1+\sin x|-\frac{1}{4\sqrt{2}}log|1-\sqrt{2}\sin x|+\frac{1}{4\sqrt{2}}log|1+\sqrt{2}\sin x|+c$
$I=\frac{1}{8}\log |\frac{\sin x-1}{\sin x+1}|-\frac{1}{4\sqrt{2}}\log |\frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1}|+c$