Question

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

Answer 1

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

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

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

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

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

Let $cosx=t$

$-sinxdx=dt$

$sinxdx=-dt$

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

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

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

Now,

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

Solving this, we get,

$A=-\frac{1}{6},B=-\frac{1}{2},C=-\frac{4}{3}$

Therefore,

$I=\int \frac{A}{t+1}dt+\int \frac{B}{t-1}dt+\int \frac{C}{1-2t}dt$

$I=Alog(t+1)+Blog(t-1)-\frac{C}{2}log(1-2t)+C$

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