Question

Integrate:

$I=\int \frac{\sin x}{\sin 4x}dx$

Answer 1

$I=\int \frac{\sin x}{\sin 4x}dx=\int \frac{\sin x}{4\sin x\cos x\cos 2x}dx=\frac{1}{4}\int \frac{dx}{\cos x(1-2{\sin }^{2}x)}$

$\Rightarrow I=\frac{1}{4}\int \frac{\cos xdx}{\cos {}^{2}x(1-2{\sin }^{2}x)}=\frac{1}{4}\int \frac{\cos xdx}{(1-2{\sin }^{2}x)(1-2{\sin }^{2}x)}$

Let $\sin x=t$

$\Rightarrow \cos xdx=dt$

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

$\Rightarrow I=\frac{1}{4}[\int \{\frac{1}{(t^2-1)}-\frac{2}{({2t}^2-1)}\}dt]=\frac{1}{4}\int \frac{1}{t^2-1}dt-\frac{1}{4}\int \frac{1}{t^2-\frac{1}{2}}dt$

$\Rightarrow I=\frac{1}{4\times 2}\iota \eta \left(\frac{t-1}{t+1}\right)-\frac{1}{4}\frac{1}{2\times \sqrt{2}}\iota \eta \left|\frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}}\right|+c$

$\Rightarrow I=\frac{1}{8}\iota \eta \left|\frac{\sin x-1}{\sin x+1}\right|-\frac{1}{4\sqrt{2}}\iota \eta \left|\frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1}\right|+c$

Hence, solve.