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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.

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watching :)
hahah ....
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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Other answers:

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

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