Evaluate $\displaystyle \int \dfrac{dx}{\sin x - \sin 2x}$
$\begin{aligned}
\text{Let } I & = \int \dfrac{dx}{\sin x - \sin 2x} \\\\
& = \int \dfrac{dx}{\sin x - 2 \sin x \cos x} \\\\
& = \int \dfrac{dx}{\sin x \left(1 - 2 \cos x\right)} \\\\
& = \int \dfrac{\sin x \; dx}{\sin^2 x \left(1 - 2 \cos x\right)} \\\\
& = \int \dfrac{\sin x \; dx}{\left(1 - \cos^2 x\right) \left(1 - 2 \cos x\right)} \\\\
& = \int \dfrac{\sin x \; dx}{\left(1 + \cos x\right) \left(1 - \cos x\right) \left(1 - 2 \cos x\right)} \;\;\; \cdots \; (1)
\end{aligned}$
Let $\cos x = u$ $\;\;\; \cdots (2a)$
Differentiating equation $(2a)$ gives
$- \sin x \; dx = du$ $\implies$ $\sin x \; dx = - du$ $\;\;\; \cdots \; (2b)$
In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$I = \displaystyle \int \dfrac{- du}{\left(1 + u\right) \left(1 - u\right) \left(1 - 2u\right)}$ $\;\;\; \cdots \; (3)$
Let $\dfrac{-1}{\left(1 + u\right) \left(1 - u\right) \left(1 - 2 u\right)} = \dfrac{A}{1 + u} + \dfrac{B}{1 - u} + \dfrac{C}{1 - 2 u}$ $\;\;\; \cdots \; (4)$
$\implies$ $-1 = A \left(1 - u\right) \left(1 - 2u\right) + B \left(1 + u\right)\left(1 - 2u\right) + C \left(1 + u\right) \left(1 - u\right)$
$\begin{aligned}
\text{When } u = -1, \hspace{2em} & -1 = A \times 2 \times 3 & \text{i.e. } -1 = 6 A & \implies A = \dfrac{-1}{6} \\\\
\text{When } u = 1, \hspace{2em} & -1 = B \times 2 \times \left(-1\right) & \text{i.e. } 1 = 2 B & \implies B = \dfrac{1}{2} \\\\
\text{When } u = \dfrac{1}{2}, \hspace{2em} & -1 = C \times \dfrac{3}{2} \times \dfrac{1}{2} & \text{i.e. } -1 = \dfrac{3 \; C}{4} & \implies C = \dfrac{-4}{3}
\end{aligned}$
Substituting the values of A, B and C in equation $(4)$ gives
$\dfrac{-1}{\left(1 + u\right) \left(1 - u\right) \left(1 - 2 u\right)} = \dfrac{-1}{6 \left(1 + u\right)} + \dfrac{1}{2 \left(1 - u\right)} + \dfrac{-4}{3 \left(1 - 2 u\right)}$ $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ In view of equation $(5)$, equation $(3)$ becomes
$\begin{aligned}
I & = \dfrac{-1}{6} \int \dfrac{du}{1 + u} + \dfrac{1}{2} \int \dfrac{du}{1 - u} - \dfrac{4}{3} \int \dfrac{du}{1 - 2 u} \\\\
& = \dfrac{-1}{6} \log \left|1 + u\right| - \dfrac{1}{2} \log \left|1 - u\right| + \dfrac{2}{3} \log \left|1 - 2 u\right| + c \\\\
& = \dfrac{-1}{6} \log \left|1 + \cos x\right| - \dfrac{1}{2} \log \left|1 - \cos x\right| + \dfrac{2}{3} \log \left|1 - 2 \cos x\right| + c \\\\
& \;\;\; \left[\text{from equation } (2a)\right]
\end{aligned}$