Evaluate: ∫ (tanx/4)/(1 - sinx/4) dx.

$\int\frac{\tan\frac{x}{4}}{1 - \sin\frac{x}{4}}dx = \int\frac{\sin\frac{x}{4}\cos\frac{x}{4}\,dx}{\left(1 - \sin\frac{x}{4}\right)\left(1 - \sin^2\frac{x}{4}\right)}$ Let $\sin\frac{x}{4} = t$, so $\cos\frac{x}{4}\,dx = 4\,dt$. $4\int\frac{t\,dt}{(1+t)(1-t)^2}$, and by partial fractions $\frac{4t}{(1+t)(1-t)^2} = \frac{-1}{1+t} + \frac{-1}{1-t} + \frac{2}{(1-t)^2}$ $= -\log(1+t) + \log(1-t) + \frac{2}{1-t} + C$ $= -\log\left(1 + \sin\frac{x}{4}\right) + \log\left(1 - \sin\frac{x}{4}\right) + \frac{2}{1 - \sin\frac{x}{4}} + C$.