Question

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.

Answer 1

watching :)

Answer 2

hahah ....

Answer 3

I think I'm just gonna watch and learn as I'm trying to eat breakfast and this technique is brand new to me :)

Answer 4

me too ... just for the sake of memory :)

Answer 5

Me three!

Answer 6

This is why OS is awesome, they never teach you subs like this in the USA...

Answer 7

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.

Answer 8

this looks promising
\[ x = a \; \sec ^2 t - b\; \tan ^2 t\]

Answer 9

looks like Euler's subs is like Weierstrass subs ... always gives the answer but really last resort.

Answer 10

Can I try the first one?

Answer 11

yep sure ...

Answer 12

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...

Answer 13

pretty cool technique ... i hadn't seen this either.

Answer 14

you see how you change dx?

Answer 15

dx = dt

Answer 16

I get it.

Answer 17

nice use of symmetry on the previous solution BTW

Answer 18

But there is a question how to find the real solution? How \(\sinh 2u\) can be found from \(t=\cosh u\)?

Answer 19

that is just inverse of cosine hyperbolic ...

Answer 20

\[ \cosh^{-1}\left( t \over c \right) = u \]

Answer 21

this should be equal to
\[ \log \left( {t \over c} + \sqrt{\left( t \over c\right)^2 - 1 } \right ) \]

Answer 22

Ok. But I don't like \(\sinh(2\;\text{arccosh}\; t) \)

Answer 23

i don't like hyperbolic functions either. because they look like circle in complex plane :)

Answer 24

But this will make easier. Euler subs will make the integral too complicated.

Answer 25

yeah ... agree on that point. I don't like Weierstrass subs either.

Answer 26

I calculated this in Wolfram Mathematica and found that it is not so similar as you have said in the beginning!

Answer 27

What is your method to find the first integral? Do you have any ideas how to calculate the second one?

Answer 28

i would try makings
\[ x = a \; \sec ^2 t - b\; \tan ^2 t \]

haven't thought of second one.

Answer 29

Open this link and cry.
http://www.wolframalpha.com/input/?i=Integrate [Sqrt[%28x^2+-+a%29*%28x^2+-+b%29]%2C+x]

Answer 30

something is really bad with chrome

Answer 31

damn ... lol. Didn't know it would get involved with two elliptical types.