Computing $\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}$ 
$$
\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}
$$
how? $x^2/\sqrt{8x-x^2-15}$ and what to do then? 

$$
-(x^2 - 8x + 15) = 1-(x^2 - 8x + 16) = 1 -(x-4)^2 = 1-\sin^2\theta
$$
So the square root of this is $\cos\theta$.
$$
dx = \cos\theta\,d\theta
$$
$$
x^2 = (4+\sin\theta)^2
$$
As $x$ goes from $3$ to $5$, $\sin\theta=x-4$ goes from $-1$ to $1$, so $\theta$ goes from $-\pi/2$ to $\pi/2$.  We get
$$
\int_{-\pi/2}^{\pi/2} \frac{(4+\sin\theta)^2}{\cos\theta}\,d\theta = \int_{-\pi/2}^{\pi/2} \frac{(4+\sin\theta)^2}{1-\sin^2\theta}\,(\cos\theta\,d\theta) = \int_{-1}^1 \frac{(4+u)^2}{1-u^2} \, du.
$$
Then partial fractions should do it.
Or you might try a rationalizing substitution rather than a trigonometric substitution.
