Evaluate: ∫ (1 + sin x)/(1 - sin x) dx

Multiply numerator and denominator by $1+\sin x$: $\frac{1+\sin x}{1-\sin x} = \frac{(1+\sin x)^2}{\cos^2 x} = (\sec x+\tan x)^2 = 2\sec^2 x - 1 + 2\sec x\tan x$ $\int (2\sec^2 x - 1 + 2\sec x\tan x)\,dx = 2\tan x - x + 2\sec x + C$ Hence the integral is $2\tan x + 2\sec x - x + C$.