Question

Integrate $\int \frac{1}{\sin x-\sin 2x}dx$

Answer 1

$\int \frac{1}{sinx-sin2x}dx$

$\int \frac{1}{sinx-2sinxcosx}dx$

$\int \frac{1}{sinx(1-2cosx)}dx$

Multiplying and dividing by $sinx$

$\int \frac{sinx}{si{n}^{2}x(1-2cosx)}dx$

$\int \frac{sinx}{(1-co{s}^{2}x)(1-2cosx)}dx$

$take$ $cosx=t,$

$-sinxdx=dt$

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

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

$\frac{-1}{(1-t)(1+t)(1-2t)}$=$\frac{A}{(1-t)}$$+$$\frac{B}{(1+t)}$+$\frac{C}{(1-2t)}$

$-1=A(1+t)(1-2t)+B(1-t)(1-2t)+C(1-t)(1+t)$

$Putting$ $t=-1$

$B=\frac{-1}{6}$

$Putting$ $t=1$

$A=\frac{1}{2}$

$Putting$ $t=\frac{1}{2}$

$C=\frac{-4}{3}$

$I$=$\frac{-1}{2}\int \frac{dt}{(1-t)}$$+$$\frac{-1}{6}\int \frac{dt}{(1+t)}$+$\frac{-4}{3}\int \frac{1}{2t-1}$

$I$=$\frac{-1}{2}log\mid (1-t)\mid$+$\frac{1}{6}log\mid (1+t)\mid$+$\frac{4}{3X2}log\mid (2x+1)\mid$

$I$=$\frac{-1}{2}log\mid (1-cosx)\mid +\frac{-1}{6}log\mid (1+cosx)\mid +\frac{-4}{3}log(1-2cosx)+C$