An integration to first order 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.
 A: The supposed answer is incorrect. Up to first order, we have $$\int_0^\pi\frac{\cos\theta\cos\left(\omega t-\frac\omega c\sqrt{a^2+r^2-2rb\cos\theta}\right)}{a\sqrt{a^2+r^2-2rb\cos\theta}}\,d\theta\approx\frac{\pi rb\omega\sin(\omega a/c-\omega t)}{2ca^2}.$$
Proof: Write $\sqrt{a^2+r^2-2rb\cos\theta}=a\sqrt{1+(r^2-2rb\cos\theta)/a^2}$, which allows us to use $$(1+x)^{\pm1/2}=1\pm\frac{x^2}2+\mathcal O(x^4)$$ for $|x|<1$ (since $a\gg r$). Then\begin{align}I&=\int_0^\pi\frac{\cos\theta\cos\left(\omega t-\frac\omega c\cdot a\sqrt{1+(r^2-2rb\cos\theta)/a^2}\right)}{a\sqrt{1+(r^2-2rb\cos\theta)/a^2}}\,d\theta\\&=\int_0^\pi\left(\frac1a+\mathcal O\left(\frac1{a^3}\right)\right)\cos\theta\cos\left(\omega t-\frac\omega c\left(a+\frac{r^2-2rb\cos\theta}{2a}+\mathcal O\left(\frac1{a^3}\right)\right)\right)\,d\theta\\&\approx\int_0^\pi\frac{\cos\theta}a\cos\left(\omega t-\frac\omega c\left(a+\frac{r^2-2rb\cos\theta}{2a}\right)\right)\,d\theta\end{align} up to first order in terms of $1/a$. The cosine addition identity yields \begin{align}I&\approx\small\int_0^\pi\frac{\cos\theta}a\left[\cos\left(\omega t-\frac\omega ca\right)\cos\left(\frac\omega c\frac{r^2-2rb\cos\theta}{2a}\right)+\sin\left(\omega t-\frac\omega ca\right)\sin\left(\frac\omega c\frac{r^2-2rb\cos\theta}{2a}\right)\right]\\&\approx\int_0^\pi\frac{\cos\theta}a\left[\cos\left(\omega t-\frac\omega ca\right)+\sin\left(\omega t-\frac\omega ca\right)\cdot\frac\omega c\frac{r^2-2rb\cos\theta}{2a}\right]\,d\theta\end{align} again up to first order, since $\cos x=1+\mathcal O(x^2)$ and $\sin x=x+\mathcal O(x^3)$. Thus, \begin{align}I&\approx\frac{\omega\sin(\omega t-\omega a/c)}{2ca^2}\int_0^\pi(r^2-2rb\cos\theta)\cos\theta\,d\theta=-\frac{\pi rb\omega\sin(\omega t-\omega a/c)}{2ca^2},\end{align} which fits reasonably well when $a\gg r$.
