I haven't been able to figure out how to do this definite integral, and I know it has a known answer. Can anyone help me work out the details for this?
The answer (and solution) are given by
$$\int_u^1\frac{\cos^{-1}x\,\mathrm dx}{\sqrt{x^2-u^2}}\stackrel?=-\frac\pi2\ln u$$
but I can't figure out a way to tackle this with the usual methods (integration by parts, residue theorem, etc.). The closest thing I've found to a clue is that (a) $\cos^{-1}x=\frac\pi 2-\sin^{-1}x$, and (b), a simpler form of the integral gives
$$\int\frac{\mathrm dx}{\sqrt{x^2-u^2}}=\log(x+\sqrt{x^2-u^2})$$
which when given the limits above becomes a sum of the integrals the inverse trigonometric functions, but I haven't been able to figure out where one would go from there.
Also, in case you're thinking about doing this as a contour integral, remember that the pole at $z=u$ is not simple...
