Question

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

Answer 1

$\int \frac{\sin x}{\sin 4x}dx$ $=\int \frac{sinx}{2.\left(2sinxcosx\right)cos2x}dx$

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

$=\frac{1}{4}\int \frac{cosxdx}{(1-si{n}^{2}x)(1-2si{n}^{2}x)}$

Substitute $=sinx=t\Rightarrow cosxdx=dt$

So, $\int \frac{sinx}{sin4x}dx=\frac{1}{4}\int \frac{dt}{(1-t^2)(1-2t^2)}$

Let $t^2=y,\frac{1}{(1-y)(1-2y)}=\frac{A}{1-y}+\frac{B}{1-2y}$

$\Rightarrow 1=A(1-2y)+B(1-y)$

Put $y=1\Rightarrow 1=A(-1)+0\Rightarrow A=-1$

$y=\frac{1}{2}\Rightarrow 1=B\times \frac{1}{2}\Rightarrow B=2$

So, $\frac{1}{(1-t^2)(1-2t^2)}=\frac{-1}{1-t^2}+\frac{2}{1-2t^2}$

$\Rightarrow \int \frac{sinx}{sin4x}dx=\frac{1}{4}\int [\frac{-1}{1-t^2}+\frac{2}{1-2t^2}]dt$

$=\frac{-1}{4}\int \frac{dt}{1-t^2}+\frac{1}{2}\int \frac{1}{1-(\sqrt{2}t{)}^2}dt$

$=\frac{-1}{4}.\frac{1}{2.1}log|\frac{1+t}{1-t}+\frac{1}{2}.\frac{1}{2\times \sqrt{2}}log|\frac{1+\sqrt{2}t}{1-\sqrt{2}t}|+c$

$(\because \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}log|\frac{a+x}{a-x}|+c)$

$=\frac{1}{8}log|\frac{1-sinx}{1+sinx}|+\frac{1}{4\sqrt{2}}log|\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}|+c$