How to evaluate the following integral$\int_a^{pa}\frac{ax}{\sqrt{(a-x)(x-pa)}}dx$? Can anyone help me to evaluate the definite integral $\int_a^{pa}\frac{ax}{\sqrt{(a-x)(x-pa)}}dx$?I encountered this integral while doing a problem of particle dynamics in Ganguly Saha(Applied Mathematics).Can this integral be evaluated without the substitution of $x=asin^2\theta+pacos^2\theta$.Please anyone suggest some other method i.e. some direct method to calculate this integral.

Let 
$I = \displaystyle \int_a^{pa} \frac{ax dx}{\sqrt{(a-x)(x-pa)}}$
Let ${(a-x)(x-pa)} = z$
Then $dz = -(2x-(p+1)a) dx$
Let's find $\lambda$ and $\mu$ such that the numerator of the integral $ax$ may be expressed as
$\lambda(-(2x-(p+1)a)) + \mu$
Comparing the coefficients, we have $\displaystyle \lambda = -\frac{a}{2}$ and $\displaystyle \mu = \frac{(p+1)a^2}{2}$
Hence we can write $I = I_1 + \displaystyle \frac{(p+1)a^2}{2}I_2$ where
$I_1 = \displaystyle -\frac{a}{2} \int_0^0 \frac{dz}{\sqrt{z}} = 0$
and 
$I_2 = \displaystyle \int_a^{pa} \frac{dx}{\sqrt{(a-x)(x-pa)}}$
Observe that the denominator of $I_2$ may be written as $\sqrt{t^2 - y^2}$ where 
$t = \displaystyle \frac{a(p-1)}{2}$ and
$y = \displaystyle x - \frac{a(p+1)}{2}$
and the integral reduces to
$\displaystyle \arcsin \frac{y}{t}$
Next, note that when $x = pa$ then $y = t$ and when $x = a$, then $y = -t$
Finally $\displaystyle I_2 = \pi$
and
$\displaystyle I = \frac{(p+1)a^2}{2}I_2 = \frac{(p+1)a^2}{2} \pi$
