For the second, here's what I got:

$$\Large{\int_1^3 \frac{dx}{\sqrt{(x-1)(3-x)}}}$$

$$\Large{\int_1^3 \frac{dx}{\sqrt{-x^2+4x-3}}}$$

$$\Large{\int_1^3 \frac{dx}{\sqrt{1 - (x-2)^2}}}$$

With \(u\) substitution \(u = x-2\), we get \(du = dx\), giving us

$$\Large{\int_1^3 \frac{du}{\sqrt{1 - u^2}}}$$

Which has a standard antiderivitive of \(\arcsin(u)\), giving us \(\arcsin(x-2)\) from 1 to 3. This gives \(\arcsin(1) - \arcsin(-1) = \pi/2 - (-\pi/2) = \pi\).

Pretty neat!

Edited June 2nd, 2017
by EN