Evaluate: ∫ (dx)/(sin x + sin 2x)

$\sin x + \sin 2x = \sin x(1 + 2\cos x)$ $I = \int \frac{\sin x\,dx}{\sin^2 x(1+2\cos x)} = \int \frac{\sin x\,dx}{(1-\cos^2 x)(1+2\cos x)}$ Put $\cos x = t$, $\sin x\,dx = -dt$: $I = -\int \frac{dt}{(1-t)(1+t)(1+2t)}$ Partial fractions: $\frac{1}{(1-t)(1+t)(1+2t)} = \frac{1/6}{1-t} - \frac{1/2}{1+t} + \frac{4/3}{1+2t}$ $\int(\ldots)dt = -\frac16\ln|1-t| - \frac12\ln|1+t| + \frac23\ln|1+2t|$ Hence $I = \frac16\ln|1-\cos x| + \frac12\ln|1+\cos x| - \frac23\ln|1+2\cos x| + C$.