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.

With the variable change $u=x/a-1$ and the shorthand $q=p-1$,
$$I=\int_a^{pa}\frac{ax}{\sqrt{(a-x)(x-pa)}}dx=a^2\int_0^{q}\frac{u+1}{\sqrt{u(q-u)}}du=a^2(I_1+I_2)$$
where, $I_1$ and $I_2$ are given below, integrated with the convenient substitution $u=q\sin^2\theta$.
$$I_1=\int_0^{q}\sqrt{\frac{u}{q-u}}du=q\int_0^{\pi/2}2\sin^2\theta d\theta=\frac{\pi}{2}q$$
$$I_2=\int_0^{q}\frac{1}{\sqrt{u(q-u)}}du=\int_0^{\pi/2}2\theta =\pi$$
Thus,
$$I=a^2\left(\frac{\pi}{2}q+\pi\right)=\frac{\pi}{2}(1+p)a^2$$
