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? 
 A: Setting $x-4=t$ gives you
$$\int_{-1}^{1}\frac{(t+4)^2}{\sqrt{1-t^2}}dt.$$
Now setting $t=\sin\theta$ gives you
$$\int_{-\pi/2}^{\pi/2}(\sin^2\theta+8\sin\theta+16)\,d\theta.$$
A: Set $x=3+2\sin^2(t)$. We then obtain
$$\int_3^5 \dfrac{x^2dx}{\sqrt{(x-3)(5-x)}} = \int_0^{\pi/2} \dfrac{2(3+2\sin^2(t))\sin(2t)dt}{\sqrt{(2\sin^2(t))\cdot(2\cos^2(t))}} = \int_0^{\pi/2}2(3+2\sin^2(t))dt$$
I trust you can finish from here.
A: $$
-(x^2 - 8x + 15) = 1-(x^2 - 8x + 16) = 1 -(x-4)^2 = 1-\sin^2\theta
$$
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.
