Question

$\int \frac{\sin x}{\sin 4x}dx=-\frac{1}{k}\log \frac{1+\sin x}{1-\sin x}+\frac{1}{4\sqrt{2}}\log \frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}$. Find the value of $k$.

Answer 1

Let $I=\int \frac{\sin x}{2\sin 2x\cos 2x}dx=\int \frac{dx}{4\cos x\cos 2x}=\int \frac{\cos xdx}{4(1-{\sin }^{2}x)(1-2{\sin }^{2}x)}$
Put $\sin x=t\Rightarrow \cos xdx=dt$
$I=\int \frac{dt}{4(1-t^2)(1-2{\sin }^{2}x)}=\int \frac{dt}{4(1-t^2)(1-2t^2)}=\frac{1}{4}\int [\frac{2}{1-2t^2}-\frac{1}{1-t^2}]dt$
$=\frac{1}{2}.\frac{1}{2\sqrt{2}}\log \frac{1+\sqrt{2}t}{1-\sqrt{2}t}-\frac{1}{4.2}\log \frac{1+t}{1-t}$
$[\because \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\log \frac{a+x}{a-x}]$