Problem with integration

May 2010
4
0
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.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
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 \(\displaystyle \frac{1}{\sqrt{b-a}} \tan^{-1}\left( \frac{x \sqrt{b-a}}{\sqrt{ 1 - bx^2} } \right) + C \)
 
Last edited:
May 2010
4
0
b-a>0

Thanks!
 
Last edited:
May 2010
4
0
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.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
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)
 
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Reactions: CyBeaR
May 2010
4
0
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.