## experimentX 2 years ago Just a similar Q. Integrate with subs:- $1. \int \sqrt{(x-a)(x-b)} \; dx$ $\cancel{ 2.\int \sqrt{(x^2-a)(x^2-b)} \; dx}$ don't try .. 2. does is not closed in elementary functions.

1. mukushla

watching :)

2. experimentX

hahah ....

3. TuringTest

I think I'm just gonna watch and learn as I'm trying to eat breakfast and this technique is brand new to me :)

4. experimentX

me too ... just for the sake of memory :)

5. Callisto

Me three!

6. TuringTest

This is why OS is awesome, they never teach you subs like this in the USA...

7. experimentX

there seems to be a substitution similar to Weierstrass subs ... called Euler's subs (not surprised to hear Euler) http://en.wikipedia.org/wiki/Euler_substitution let's see if we can find something any better.

8. experimentX

this looks promising $x = a \; \sec ^2 t - b\; \tan ^2 t$

9. experimentX

looks like Euler's subs is like Weierstrass subs ... always gives the answer but really last resort.

10. klimenkov

Can I try the first one?

11. experimentX

yep sure ...

12. klimenkov

1. Using the sub $$x=t+\frac{a+b}2$$, so $$\int \sqrt{(x-a)(x-b)} \; dx=\int \sqrt{(t-\frac{a-b}2)(t+\frac{a-b}2)}\;dt=\int\sqrt{t^2-\frac{(a-b)^2}4}\;dt$$ Now let $$\frac{a-b}2=c$$ use $$t=c\cosh u,\; dt=c\sinh udu$$: $$\int\sqrt{t^2-c^2}\;dt=\int c^2\sinh^2u\;du =\frac{c^2}{2}(\frac12\sinh2u-u)+C$$ Now I must think how to make an inverse substitution...

13. experimentX

pretty cool technique ... i hadn't seen this either.

14. Zekarias

you see how you change dx?

15. experimentX

dx = dt

16. Zekarias

I get it.

17. experimentX

nice use of symmetry on the previous solution BTW

18. klimenkov

But there is a question how to find the real solution? How $$\sinh 2u$$ can be found from $$t=\cosh u$$?

19. experimentX

that is just inverse of cosine hyperbolic ...

20. experimentX

$\cosh^{-1}\left( t \over c \right) = u$

21. experimentX

this should be equal to $\log \left( {t \over c} + \sqrt{\left( t \over c\right)^2 - 1 } \right )$

22. klimenkov

Ok. But I don't like $$\sinh(2\;\text{arccosh}\; t)$$

23. experimentX

i don't like hyperbolic functions either. because they look like circle in complex plane :)

24. klimenkov

But this will make easier. Euler subs will make the integral too complicated.

25. experimentX

yeah ... agree on that point. I don't like Weierstrass subs either.

26. klimenkov

I calculated this in Wolfram Mathematica and found that it is not so similar as you have said in the beginning!

27. klimenkov

What is your method to find the first integral? Do you have any ideas how to calculate the second one?

28. experimentX

i would try makings $x = a \; \sec ^2 t - b\; \tan ^2 t$ haven't thought of second one.

29. klimenkov

Open this link and cry. http://www.wolframalpha.com/input/?i=Integrate[Sqrt[%28x^2+-+a%29*%28x^2+-+b%29]%2C+x]

30. experimentX

something is really bad with chrome

31. experimentX

damn ... lol. Didn't know it would get involved with two elliptical types.