# Math Help - Problem with integration

1. ## Problem with integration

Hi. How do you integrate this function?
$\int_0^1 \left(\alpha^2 -\beta^2 x^2\right)^{-1/2} \left(\gamma^2 -k^2 \beta^2 x^2 \right)^{-1} dx$
where:
$\alpha, \gamma > 0$ , $0 \leq k \leq 1$ and $0\leq \beta \leq 2k$
are some constants.

2. Originally Posted by CyBeaR
Hi. How do you integrate this function?
$\int_0^1 \left(\alpha^2 -\beta^2 x^2\right)^{-1/2} \left(\gamma^2 -k^2 \beta^2 x^2 \right)^{-1} dx$
where:
$\alpha, \gamma > 0$ , $0 \leq k \leq 1$ and $0\leq \beta \leq 2k$
are some constants.

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

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

Sub $x = \frac{1}{t}$

$dx = -\frac{dt}{t^2}$

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

$= - \int \frac{ t}{ (t^2 - a) \sqrt{t^2 -b} }
~dt$

Sub. $t^2 -b = u^2$

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

$= -\int \frac{du}{ u^2 + b-a}$

What do you think ? $b-a > 0$ or $< 0$

$b-a > 0$ . Let's continue

$= -\int \frac{du}{ u^2 + \sqrt{b-a}^2 }$

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

$u = \sqrt{t^2 - b} = \frac{ \sqrt{ 1 - bx^2} }{x}$
so

the integral is

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

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

3. b-a>0

Thanks!

4. How do You obtain the 2nd solution?
I've tried to calculate it without any luck. I've found in "Tables of indefinite integrals." [Petit Bois G.,p.50] that the only solution for (b-a)>0 is the later one.
I'd like to understand what's behind this.

5. Originally Posted by CyBeaR
How do You obtain the 2nd solution?
I've tried to calculate it without any luck. I've found in "Tables of indefinite integrals." [Petit Bois G.,p.50] that the only solution for (b-a)>0 is the later one.
I'd like to understand what's behind this.
Do you mean you are confused why i put 'or' to connect them ?

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

Actually , the second one can be deduced from the first one .

because we have $\tan( \frac{\pi}{2} - x ) = \frac{1}{\tan{x}}$

so $\frac{\pi}{2} - x = \tan^{-1}( \frac{1}{\tan{x}} )$

Let $t = \tan{x}$ and we have $\frac{\pi}{2} - \tan^{-1}(t) = \tan^{-1}(\frac{1}{t})$

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

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

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

I use $C$ again instead of $C -\frac{\pi}{2} \frac{1}{\sqrt{b-a}}$

6. I didn't have relation: $\tan( \frac{\pi}{2} - x ) = \frac{1}{\tan{x}}$ within arm's reach and I've tried other ones.
Now, all is clear to me. Thank You very much.