Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having some trouble evaluating an integral -- involving taking an approximation. It would be great if someone could help me.

I wish to evaluate $$\int_0^\pi {\cos\theta\cos \left[\omega t-{\omega \over c}(a^2+r^2-2rb\cos \theta)^{1\over 2}\right]\over (a^2+r^2-2rb\cos \theta)^{1\over 2}}d\theta$$

for the case where $a\gg r$ and $b=a\sin \phi$, to first order in $1\over a$.

The answer is supposed to be $$\pi r^2\omega(\sin \phi)[\sin(\omega t-{\omega\over c}a)]\over 2a$$but I can't see how to get there...

There must be some Taylor expansion required. I suppose we could write $$(a^2+r^2-2rb\cos \theta)^{1\over 2}\approx a\left(1-{rb\over a^2}\cos\theta\right)$$

but when next?

Ok, if we ignore the given answer -- assuming it is wrong, what should I be getting?

For the argument of the $\cos \left[\omega t-{\omega \over c}(a^2+r^2-2rb\cos \theta)^{1\over 2}\right]$, suppose I Taylor expand $\cos$. It would go something like $$1-{1\over 2}\left[\omega t-{\omega \over c}(a^2+r^2-2rb\cos \theta)^{1\over 2}\right]^2+...$$ However, note that the higher order terms involve terms of $\left(1\over a\right)^n$, with values of $n$ of the non-negligible magnitude. So that can't be the way to do it?

Please help! Thanks.

share|cite|improve this question
You're doing it right. Taylor-expand in $1/a$ and you'll get a simpler integral. Are you having trouble with this step of expansion? Or do you have the integral already and are having trouble evaluating it? – Marek Jan 16 '13 at 14:42
@Marek: Thank you for replying!! Yes, I am having trouble evaluating that already. There is a $\cos\theta$ in a $\cos$... – user58388 Jan 16 '13 at 22:04
@Marek: I guess I can Taylor expand that outer cos with argument containing the $\cos \theta$. But then every term in the expanded series will contain terms of non-negligible orders in $1\over a$, so it doesn't quite work? – user58388 Jan 16 '13 at 22:49
Well, for integrals like $\cos (A+ B\cos \theta)$ you can try substitutions of the form $x = A + B\cos \theta$. This will also get you rid of the other $\cos \theta$ factors. But looking at the answer you're supposed to obtain again, it seems very weird. First of all, what is $\phi$ doing there? Can you please double-check all the expressions you've entered to make sure we are trying to solve the right problem? – Marek Jan 17 '13 at 9:02
@Marek:Thanks! Yes, I have checked. It is correct. The $\phi$ comes from $b=a\sin \phi$. – user58388 Jan 17 '13 at 12:14

Your Answer


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

Browse other questions tagged or ask your own question.