Question

$Find\int \frac{\sin x}{\sin 4x}dx$

Answer 1

$\int \frac{\sin x}{\sin 4x}dx=\int \frac{\sin x}{2\sin 2x\cos 2x}=\int \frac{\sin x}{4\sin x\cos x\cos 2x}dx$

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

If we multiply by $\cos x$ in num. dena and convert ${\cos }^{2}x$ if $1-{\sin }^{2}x$ and then let $\sin x$ and then proceed with partial fraction.

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

Let $\sin x=t\Rightarrow \cos xdx=dt$

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

from this use of only its complicated use of many variable

two variable when we

put $t^2=\alpha$

$\frac{1}{(1-t^2)(1-2t^2)}\to put{t}^2=\alpha \Rightarrow \frac{1}{(1-\alpha )(1-2\alpha )}$

$\frac{1}{(1-\alpha )(1-2\alpha )}=\frac{A}{1-\alpha }+\frac{\beta }{1-2\alpha }=\frac{A(1-2\alpha )+B(1-\alpha )}{(1-\alpha )(1-2\alpha )}$

on comparison

$1=A(1-2\alpha )+B(1-\alpha )$

put $\alpha =1$

$1=A