Compute $\int_{\alpha}^{\beta} \frac{x}{\sqrt{(x-\alpha)(\beta-x)}} \ dx$ 
Possible Duplicate:
Computing $ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$ 

How would you compute 
$$\int_{\alpha}^{\beta} \frac{x}{\sqrt{(x-\alpha)(\beta-x)}} \ dx\space?$$
 A: Make a change of variables $y= \alpha+\beta - x$.
$$
  \mathcal{I}_{\alpha,\beta} := \int_\alpha^\beta \frac{x}{\sqrt{(x-\alpha)(\beta-x)}} \mathrm{d}x = \int_{\alpha}^\beta \frac{\alpha+\beta - y}{\sqrt{(\beta-y)(y-\alpha)}} \mathrm{d}y
$$
Implying
$$
    \frac{2}{\alpha+\beta} \mathcal{I}_{\alpha,\beta} =  \int_\alpha^\beta \frac{\mathrm{d}x}{\sqrt{(x-\alpha)(\beta-x)}}
$$
Now standardize: $x = \alpha + (\beta-\alpha) u$:
$$
    \frac{2}{\alpha+\beta} \mathcal{I}_{\alpha,\beta} = \int_0^1 \frac{\mathrm{d} u}{\sqrt{u(1-u)}}
$$
Giving:
$$
      \mathcal{I}_{\alpha,\beta} = \frac{\alpha+\beta}{2} \int_0^1 \frac{\mathrm{d} u}{\sqrt{u(1-u)}} = \frac{\alpha+\beta}{2} \left. 2 \arcsin(\sqrt{u}) \right|_{u=0}^{u=1} = \pi  \frac{\alpha+\beta}{2} 
$$
A: Let $$I(a,b) = \int_a^b \dfrac{x}{\sqrt{(x-a)(b-x)}} dx$$
$$I(-b,b) = \int_{-b}^b \dfrac{x}{\sqrt{(x+b)(b-x)}} dx = 0 \,\,\,\,\,\,\,(\text{Odd function})$$
Hence, $$I(a,b) = f(a+b)$$
Hence, $$f(z) = I(0,z) = \int_0^z \dfrac{\sqrt{x}}{\sqrt{z-x}} dx$$
Now $x = z-t^2 \implies dx = -2t dt$.
Hence, $$f(z) = I(0,z) = \int_0^z \dfrac{\sqrt{x}}{\sqrt{z-x}} dx = \int_{\sqrt{z}}^{0} \dfrac{\sqrt{z-t^2}}{t}(-2t dt) = 2\int_0^{\sqrt{z}} \sqrt{z-t^2} dt = 2 \dfrac{\pi z}4 = \dfrac{\pi z}2$$
Hence, $$I(a,b) = f(a+b) = \dfrac{\pi(a+b)}2$$
