Evaluate: ∫ √(csc x - 1) dx

$\sqrt{\csc x - 1} = \sqrt{\frac{1 - \sin x}{\sin x}} = \sqrt{\frac{(1 - \sin x)(1 + \sin x)}{\sin x(1 + \sin x)}} = \frac{\cos x}{\sqrt{\sin x + \sin^2 x}}$ Put $\sin x = t$, so $\cos x\, dx = dt$. $I = \int \frac{dt}{\sqrt{t^2 + t}} = \int \frac{dt}{\sqrt{\left(t + \frac12\right)^2 - \left(\frac12\right)^2}}$ $= \log\left|t + \frac12 + \sqrt{t^2 + t}\right| + C$ Hence $I = \log\left|\sin x + \frac12 + \sqrt{\sin x + \sin^2 x}\right| + C$.