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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.
 2 years ago
 2 years 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.
 2 years ago
 2 years ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
hahah ....
 2 years 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 :)
 2 years ago

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

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

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

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

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
nice use of symmetry on the previous solution BTW
 2 years 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 \(t=\cosh u\)?
 2 years ago

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

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah ... agree on that point. I don't like Weierstrass subs either.
 2 years 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!
 2 years 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?
 2 years 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.
 2 years 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]
 2 years ago

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