# Problem with integration

• May 16th 2010, 11:44 PM
CyBeaR
Problem with integration
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.
• May 17th 2010, 12:09 AM
simplependulum
Quote:

Originally Posted by CyBeaR
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 $\displaystyle \frac{1}{\sqrt{b-a}} \tan^{-1}\left( \frac{x \sqrt{b-a}}{\sqrt{ 1 - bx^2} } \right) + C$
• May 17th 2010, 12:26 AM
CyBeaR
b-a>0

Thanks!
• May 18th 2010, 05:34 AM
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.
• May 18th 2010, 08:20 PM
simplependulum
Quote:

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 ?

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

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

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

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

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

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

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

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

I use $\displaystyle C$ again instead of $\displaystyle C -\frac{\pi}{2} \frac{1}{\sqrt{b-a}}$ (Happy)
• May 18th 2010, 11:30 PM
CyBeaR
I didn't have relation: $\displaystyle \tan( \frac{\pi}{2} - x ) = \frac{1}{\tan{x}}$ within arm's reach and I've tried other ones.
Now, all is clear to me. Thank You very much.