# Problem with integration

#### 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.

#### simplependulum

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

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b-a>0

Thanks!

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#### 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.

#### simplependulum

MHF Hall of Honor
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)

CyBeaR

#### 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.