To compute improper integral $\int_3^{5}\frac{x^{2}\, dx}{\sqrt{x-3}{\sqrt{5-x}}}$ I am given improper integral as 
$$\int_3^{5}\frac{x^{2}}{\sqrt{x-3}{\sqrt{5-x}}}dx$$
DOUBT
I see that problem is at both the end points, so i need to split up the integral. But problem seems to me that when i split up integral one of terms in denominator which is ${5-x}$ becomes negative. So how do i split up the integral ? Kindly help
Thanks
 A: Beta Function Approach
Substituting $x\mapsto2x+3$,
$$
\begin{align}
&\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\
&=\int_0^1\frac{(2x+3)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=\int_0^1\frac{(3(1-x)+5x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}
+30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}}
+25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\operatorname{B}\left(\frac52,\frac12\right)
+30\operatorname{B}\left(\frac32,\frac32\right)
+25\operatorname{B}\left(\frac12,\frac52\right)\\
&=9\cdot\frac38\pi
+30\cdot\frac18\pi
+25\cdot\frac38\pi\\
&=\frac{33}2\pi
\end{align}
$$
using the Beta Function.

Trigonometric Subsitution
After the substitution $x\mapsto2x+3$, we can use $x\mapsto\sin^2(x)$
$$
\begin{align}
&\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\
&=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}
+30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}}
+25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\
&=9\int_0^{\pi/2}2\cos^4(x)\,\mathrm{d}x
+30\int_0^{\pi/2}2\sin^2(x)\cos^2(x)\,\mathrm{d}x
+25\int_0^{\pi/2}2\sin^4(x)\,\mathrm{d}x\\
&=9\cdot\frac38\pi
+30\cdot\frac18\pi
+25\cdot\frac38\pi\\
&=\frac{33}2\pi
\end{align}
$$
A: Hint:
$$(5-x)(x-3)=(-x^2-15+8x)=(1-(x-4)^2)$$
now let $x=sint$ and you will get rid of the square root. And be careful about the range, which changes when you changing your variable.
A: Notice, the following 
$$x^2=A(x-3)(5-x)+B(8-2x)+C$$
On solving we get 
$$x^2=-(x-3)(5-x)-4(8-2x)+17$$
Now, we have 
$$\int_{3}^{5}\frac{x^2dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=\int_{3}^{5}\frac{(-(x-3)(5-x)-4(8-2x)+17)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=-\int_{3}^{5}\frac{(x-3)(5-x)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}+17\int_{3}^{5}\frac{dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$
$$=-\int_{3}^{5}\sqrt{(x-3)}\sqrt{(5-x)}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{8x-x^2-15}}+17\int_{3}^{5}\frac{dx}{\sqrt{1-(x-4)^2}}$$
$$=-\int_{3}^{5}\sqrt{1-(x-4)^2}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{8x-x^2-15}}+17\int_{3}^{5}\frac{dx}{\sqrt{1-(x-4)^2}}$$
$$=-\frac{1}{2}\left[(x-4)\sqrt{1-(x-4)^2}+\sin^{-1}\left(x-4\right)\right]_{3}^{5}-4\left[2\sqrt{1-(x-4)^2}\right]_{3}^{5}+17\left[\sin^{-1}(x-4)\right]_{3}^{5}$$
$$=-\frac{1}{2}(\pi)-8(0)+17(\pi)=\frac{33\pi}{2}$$
A: I'd suggest that you do $u=x-4$ to get an integral from $-1$ to $1$. It will be
$$
\int_{-1}^1\frac{(u+4)^2}{\sqrt{1-u^2}}\,du=\int_{-1}^1-\sqrt{1-u^2}+\frac{8u}{\sqrt{1-u^2}}+\frac{17}{\sqrt{1-u^2}}\,du.
$$
The first term in the integrand gives $-\pi/2$, since it represents (minus) the area of a half circle. The second term gives nothing, since it is odd, and the interval is even. The last term gives
$$
\bigl[17\arcsin u\bigr]_{-1}^1=17(\pi/2-(-\pi/2))=17\pi.
$$
All in all, the integral equals $-\pi/2+17\pi=33\pi/2$.
