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I need to find:

$$\int \sqrt{\frac{x-a}{x+a}} ~dx$$

I want to substitute $x$ with $a*\cosh$,but I get confused ,because $x \gt a$ or $ x \lt -a$ ,and $a*\cosh$ works only in the first case. Could you please find this integral and justify each step you make, especially the part with substitution?

Do I need to show that there is one-to-one correspondence between the domain of my variable and the range of the new function?

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I apologize ,it should be x-a/x+a – user1978522 Feb 26 '13 at 21:09
Then fix it in the problem please – Ron Gordon Feb 26 '13 at 21:11
What are the limits of integration? – Christopher A. Wong Feb 26 '13 at 21:13
It's an indefinite integral – user1978522 Feb 26 '13 at 21:14
up vote 4 down vote accepted

Yes, substitute $x=a \cosh{t}$, $dx=a \sinh{t} dt$, get

$$a\int dt \sinh{t} \tanh{(t/2)} = a\int dt\: 2 \sinh^2{(t/2)}$$

or the integral equals

$$a \int dt \: (\cosh{t}-1) = a (\sinh{t}-t)+C = a \left [ \sqrt{\frac{x^2}{a^2}-1} - \mathrm{arccosh}{\left(\frac{x}{a}\right)}\right ]+C$$

where $C$ is an integration constant.

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What about the case when x<-a ? – user1978522 Feb 26 '13 at 21:24
Then you have imaginary arguments, and the cosh's become cosines. But in that case, you wouldn't express the integrand like that; you would have the $x$ and $a$ reversed. So the analysis is certainly valid, but it involves imaginary numbers, that's all. – Ron Gordon Feb 26 '13 at 21:27
How about $x = -a\cosh t$? – J.H. Feb 26 '13 at 21:28
It will produce a factor of $i$ and a $+ t$ instead of a $-t$. But then substitute back and everything should work out. Really, try this yourself. – Ron Gordon Feb 26 '13 at 21:33
Do you take t>0 and why? – user1978522 Feb 26 '13 at 21:46

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