experimentX
  • experimentX
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
watching :)
experimentX
  • experimentX
hahah ....
TuringTest
  • TuringTest
I think I'm just gonna watch and learn as I'm trying to eat breakfast and this technique is brand new to me :)

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experimentX
  • experimentX
me too ... just for the sake of memory :)
Callisto
  • Callisto
Me three!
TuringTest
  • TuringTest
This is why OS is awesome, they never teach you subs like this in the USA...
experimentX
  • experimentX
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.
experimentX
  • experimentX
this looks promising \[ x = a \; \sec ^2 t - b\; \tan ^2 t\]
experimentX
  • experimentX
looks like Euler's subs is like Weierstrass subs ... always gives the answer but really last resort.
klimenkov
  • klimenkov
Can I try the first one?
experimentX
  • experimentX
yep sure ...
klimenkov
  • klimenkov
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...
experimentX
  • experimentX
pretty cool technique ... i hadn't seen this either.
anonymous
  • anonymous
you see how you change dx?
experimentX
  • experimentX
dx = dt
anonymous
  • anonymous
I get it.
experimentX
  • experimentX
nice use of symmetry on the previous solution BTW
klimenkov
  • klimenkov
But there is a question how to find the real solution? How \(\sinh 2u\) can be found from \(t=\cosh u\)?
experimentX
  • experimentX
that is just inverse of cosine hyperbolic ...
experimentX
  • experimentX
\[ \cosh^{-1}\left( t \over c \right) = u \]
experimentX
  • experimentX
this should be equal to \[ \log \left( {t \over c} + \sqrt{\left( t \over c\right)^2 - 1 } \right ) \]
klimenkov
  • klimenkov
Ok. But I don't like \(\sinh(2\;\text{arccosh}\; t) \)
experimentX
  • experimentX
i don't like hyperbolic functions either. because they look like circle in complex plane :)
klimenkov
  • klimenkov
But this will make easier. Euler subs will make the integral too complicated.
experimentX
  • experimentX
yeah ... agree on that point. I don't like Weierstrass subs either.
klimenkov
  • klimenkov
I calculated this in Wolfram Mathematica and found that it is not so similar as you have said in the beginning!
klimenkov
  • klimenkov
What is your method to find the first integral? Do you have any ideas how to calculate the second one?
experimentX
  • experimentX
i would try makings \[ x = a \; \sec ^2 t - b\; \tan ^2 t \] haven't thought of second one.
klimenkov
  • klimenkov
Open this link and cry. http://www.wolframalpha.com/input/?i=Integrate[Sqrt[%28x^2+-+a%29*%28x^2+-+b%29]%2C+x]
experimentX
  • experimentX
something is really bad with chrome
experimentX
  • experimentX
damn ... lol. Didn't know it would get involved with two elliptical types.

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