Evaluate: ∫ tan⁻¹√(1-x)/(1+x) dx.

Put $x = \cos 2\theta$, so $\theta = \frac{1}{2}\cos^{-1}x$. $\sqrt{\frac{1 - x}{1 + x}} = \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} = \tan\theta$, so $\tan^{-1}\sqrt{\frac{1-x}{1+x}} = \theta = \frac{1}{2}\cos^{-1}x$. $\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx = \frac{1}{2}\int\cos^{-1}x\,dx$ By parts: $\int\cos^{-1}x\,dx = x\cos^{-1}x + \int\frac{x}{\sqrt{1 - x^2}}dx = x\cos^{-1}x - \sqrt{1 - x^2}$ Hence the integral is $\frac{1}{2}\left(x\cos^{-1}x - \sqrt{1 - x^2}\right) + C$.