A community for students.
Here's the question you clicked on:
 0 viewing
experimentX
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1me too ... just for the sake of memory :)

TuringTest
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1this looks promising \[ x = a \; \sec ^2 t  b\; \tan ^2 t\]

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

klimenkov
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1pretty cool technique ... i hadn't seen this either.

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

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

klimenkov
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1that is just inverse of cosine hyperbolic ...

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

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

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

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

klimenkov
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1something is really bad with chrome

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1damn ... lol. Didn't know it would get involved with two elliptical types.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.