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experimentX
Group Title
Just a similar Q.
Integrate with subs:
\[ 1. \int \sqrt{(xa)(xb)} \; dx\]
\[ \cancel{ 2.\int \sqrt{(x^2a)(x^2b)} \; dx}\]
don't try .. 2. does is not closed in elementary functions.
 one year ago
 one year ago
experimentX Group Title
Just a similar Q. Integrate with subs: \[ 1. \int \sqrt{(xa)(xb)} \; dx\] \[ \cancel{ 2.\int \sqrt{(x^2a)(x^2b)} \; dx}\] don't try .. 2. does is not closed in elementary functions.
 one year ago
 one year ago

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mukushla Group TitleBest ResponseYou've already chosen the best response.0
watching :)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
hahah ....
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I think I'm just gonna watch and learn as I'm trying to eat breakfast and this technique is brand new to me :)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
me too ... just for the sake of memory :)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Me three!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
This is why OS is awesome, they never teach you subs like this in the USA...
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this looks promising \[ x = a \; \sec ^2 t  b\; \tan ^2 t\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
looks like Euler's subs is like Weierstrass subs ... always gives the answer but really last resort.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
Can I try the first one?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yep sure ...
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
1. Using the sub \(x=t+\frac{a+b}2\), so \(\int \sqrt{(xa)(xb)} \; dx=\int \sqrt{(t\frac{ab}2)(t+\frac{ab}2)}\;dt=\int\sqrt{t^2\frac{(ab)^2}4}\;dt\) Now let \(\frac{ab}2=c\) use \(t=c\cosh u,\; dt=c\sinh udu\): \(\int\sqrt{t^2c^2}\;dt=\int c^2\sinh^2u\;du =\frac{c^2}{2}(\frac12\sinh2uu)+C\) Now I must think how to make an inverse substitution...
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
pretty cool technique ... i hadn't seen this either.
 one year ago

Zekarias Group TitleBest ResponseYou've already chosen the best response.0
you see how you change dx?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dx = dt
 one year ago

Zekarias Group TitleBest ResponseYou've already chosen the best response.0
I get it.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
nice use of symmetry on the previous solution BTW
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
But there is a question how to find the real solution? How \(\sinh 2u\) can be found from \(t=\cosh u\)?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
that is just inverse of cosine hyperbolic ...
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ \cosh^{1}\left( t \over c \right) = u \]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this should be equal to \[ \log \left( {t \over c} + \sqrt{\left( t \over c\right)^2  1 } \right ) \]
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
Ok. But I don't like \(\sinh(2\;\text{arccosh}\; t) \)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i don't like hyperbolic functions either. because they look like circle in complex plane :)
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
But this will make easier. Euler subs will make the integral too complicated.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah ... agree on that point. I don't like Weierstrass subs either.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
I calculated this in Wolfram Mathematica and found that it is not so similar as you have said in the beginning!
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
What is your method to find the first integral? Do you have any ideas how to calculate the second one?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
i would try makings \[ x = a \; \sec ^2 t  b\; \tan ^2 t \] haven't thought of second one.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.3
Open this link and cry. http://www.wolframalpha.com/input/?i=Integrate[Sqrt[%28x^2++a%29*%28x^2++b%29]%2C+x]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
something is really bad with chrome
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
damn ... lol. Didn't know it would get involved with two elliptical types.
 one year ago
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