Question

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

Answer 1

Let I = $\int \frac{dx}{sinx-sin2x}\Rightarrow I=\int \frac{dx}{sinx(1-2cosx)}\Rightarrow I=\int \frac{sinxdx}{(1+cosx)(1-cosx)(1-2cosx)}\text{Put}cosx=t\Rightarrow sinxdx=-dt\therefore I=\int \frac{-dt}{(1+t)(1-t)(1-2t)}\text{Consider}\frac{1}{(1+t)(1-t)(1-2t)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{1-2t}\Rightarrow 1=A(1-t)(1-2t)+B(1+t)(1-2t)+C(1-t)(1+t)\text{On equating the coefficients of like terms, we get:}A=\frac{1}{6},B=-\frac{1}{2},C=\frac{4}{3}So,I=-\frac{1}{6}\int \frac{dt}{(1+t)}+\frac{1}{2}\int \frac{dt}{(1-t)}-\frac{4}{3}\int \frac{dt}{(1-2t)}\Rightarrow I=-\frac{1}{6}log|1+t|-\frac{1}{2}log|1-t|+\frac{2}{3}log|1-2t|+C\therefore I=-\frac{1}{6}log|1+cosx|-\frac{1}{2}log|1-cosx|+\frac{2}{3}log|1-2cosx|+C.$