## experimentX Group Title 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. one year ago one year ago

1. mukushla Group Title

watching :)

2. experimentX Group Title

hahah ....

3. TuringTest Group Title

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 Group Title

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

5. Callisto Group Title

Me three!

6. TuringTest Group Title

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

7. experimentX Group Title

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 Group Title

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

9. experimentX Group Title

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

10. klimenkov Group Title

Can I try the first one?

11. experimentX Group Title

yep sure ...

12. klimenkov Group Title

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 Group Title

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

14. Zekarias Group Title

you see how you change dx?

15. experimentX Group Title

dx = dt

16. Zekarias Group Title

I get it.

17. experimentX Group Title

nice use of symmetry on the previous solution BTW

18. klimenkov Group Title

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

19. experimentX Group Title

that is just inverse of cosine hyperbolic ...

20. experimentX Group Title

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

21. experimentX Group Title

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

22. klimenkov Group Title

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

23. experimentX Group Title

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

24. klimenkov Group Title

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

25. experimentX Group Title

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

26. klimenkov Group Title

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

27. klimenkov Group Title

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

28. experimentX Group Title

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

29. klimenkov Group Title

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 Group Title

something is really bad with chrome

31. experimentX Group Title

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