Hi. How do you integrate this function?

\(\displaystyle \int_0^1 \left(\alpha^2 -\beta^2 x^2\right)^{-1/2} \left(\gamma^2 -k^2 \beta^2 x^2 \right)^{-1} dx\)

where:

\(\displaystyle \alpha, \gamma > 0\) , \(\displaystyle 0 \leq k \leq 1\) and \(\displaystyle 0\leq \beta \leq 2k\)

are some constants.

Let's solve the INdefinite integral , to be convenient , i would rewrite the integral

\(\displaystyle \int \frac{dx}{ (1-ax^2) \sqrt{ 1 - bx^2 } }\)

Sub \(\displaystyle x = \frac{1}{t} \)

\(\displaystyle dx = -\frac{dt}{t^2} \)

\(\displaystyle = \int \frac{t^3}{ (t^2-a) \sqrt{ t^2 - b } }\left( \frac{-dt}{t^2} \right)\)

\(\displaystyle = - \int \frac{ t}{ (t^2 - a) \sqrt{t^2 -b} }

~dt \)

Sub. \(\displaystyle t^2 -b = u^2 \)

\(\displaystyle = -\int \frac{u~du}{ (u^2 + b - a )u}\)

\(\displaystyle = -\int \frac{du}{ u^2 + b-a} \)

What do you think ? \(\displaystyle b-a > 0 \) or \(\displaystyle < 0 \)

\(\displaystyle b-a > 0 \) . Let's continue

\(\displaystyle = -\int \frac{du}{ u^2 + \sqrt{b-a}^2 } \)

\(\displaystyle = \frac{-1}{\sqrt{b-a}} \tan^{-1}\left( \frac{u}{\sqrt{b-a}} \right) + C \)

\(\displaystyle u = \sqrt{t^2 - b} = \frac{ \sqrt{ 1 - bx^2} }{x} \)

so

the integral is

\(\displaystyle \frac{-1}{\sqrt{b-a}} \tan^{-1}\left( \frac{\sqrt{ 1 - bx^2} }{x \sqrt{b-a}} \right) + C \)

or \(\displaystyle \frac{1}{\sqrt{b-a}} \tan^{-1}\left( \frac{x \sqrt{b-a}}{\sqrt{ 1 - bx^2} } \right) + C \)