Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to evaluate $ \int \frac{1}{x} \sqrt{\frac{x+a}{x-a}}dx $.

$ x=a\cosh(2t), \int \frac{1}{x} \sqrt{\frac{x+a}{x-a}}dx= \int \frac{2\tanh(2t)}{\tanh(t)}dt= \int \frac{4}{1+\tanh^2(t)}dt $

$ u=\tanh(t), \int \frac{4}{1+\tanh^2(t)}dt=2 \int (\frac{1}{1+u^2}+\frac{1}{1-u^2})du=2 \mathrm{artanh}(u)+2\arctan(u)+C= $ $ \mathrm{arcosh}(x/a)+2\arctan(\sqrt{\frac{x^2-a^2}{x^2+a^2}})+C $

However, I could not manage to show that the derivative of this function is $ \frac{1}{x} \sqrt{\frac{x+a}{x-a}} $.

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

It seems you made a mistake in expressing $u$ in terms of $x$. It's $u=\sqrt{(x-a)/(x+a)}$, so the second term should be $2\arctan\sqrt{(x-a)/(x+a)}$, without the squares. That makes the derivative come out right.

share|improve this answer
add comment

After substitutions back into the correct anti-derivative in terms of $u$, I get different result:

$$ \cosh^{-1}\left(\frac{x}{a}\right) + 2 \arctan\left( \sqrt{ \frac{x-a}{x+a} }\right) + C $$

Differentiating this, I get $\dfrac{1}{x} \left( \dfrac{x-a}{x+a} \right)^{-\frac12}$.

share|improve this answer
    
He seems to have gotten the "hyperbolic arc tangent", rather; and that is correct: the derivative of $\mathrm{artanh}(u)$ is $\frac{1}{1-u^2}$, since $\mathrm{artanh}(z) = \frac{1}{2}\ln((1+z)/(1-z))$, i.e., what you have. –  Arturo Magidin Aug 27 '11 at 19:46
    
@Arturo: Is it really called "hyperbolic arc tangent"? Since the "ar" is for "area", I thought the full name would me "(hyperbolic) area tangent"? –  joriki Aug 27 '11 at 19:49
    
@joriki: Honestly, I don't know. I don't use these functions much. Like the inverse trig functions, I'm sure they have a lot of names; but I suspect you are far more likely to be right than I am. –  Arturo Magidin Aug 27 '11 at 19:52
    
$ 2 \int \frac{1}{1-u^2}du= \log(1+u)-\log(1-u)=2artanh(u)... $ –  Chon Aug 27 '11 at 20:05
    
Thank you Arturo, I have updated my answer, the issue was in substituting back the chain of transformations from $u$ to $x$. –  Sasha Aug 27 '11 at 20:19
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.