Evaluating $\int \frac{\sin^2(x)}{1-\cos(x)} \, \text{d}x$.

Question

I need to find, $$\int\frac{\sin^2(x)}{1-\cos(x)}\,\text{d}x.$$ I tried substituting $$u = 1 - \cos (x)$$ or $$u = \cos(x)$$ and even rewriting $$\sin^2(x) = 1 - \cos^2(x)$$ But neither works.

How should I bite it?

Answer 1

Notice that $$\frac{\sin^2 x}{1 - \cos x} = \frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x}$$

Answer 2

HINT: $$\frac{\sin^2x}{1-\cos x}=1+\cos x$$