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experimentX
 2 years ago
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.
experimentX
 2 years ago
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.

This Question is Closed

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0This is why OS is awesome, they never teach you subs like this in the USA...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1there 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.

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1looks like Euler's subs is like Weierstrass subs ... always gives the answer but really last resort.

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

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.31. 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...

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1nice use of symmetry on the previous solution BTW

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

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

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

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

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

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

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.3But this will make easier. Euler subs will make the integral too complicated.

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

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

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

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

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

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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1damn ... lol. Didn't know it would get involved with two elliptical types.
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