<s>[INST] How to simplify this complex integral? How to approximate this integral as a function of a and b?
$$\int_0^\pi\int_0^{2\pi}\sqrt{(a-b\sin\varphi\cos\theta)^2+(b\cos\varphi)^2+(b\sin\varphi\sin\theta)^2}d\theta d\varphi$$
where a and b are two variables.
 [/INST] As was mentioned: the integrand simplifies to: $\sqrt{a^2 + b^2 - 2 a b \cos\theta \sin\phi}$
Factor out $\sqrt{a^2 +b^2}$ and use the binomial theorem to expand, treating $\cos\theta \sin \phi$ as small. This gives:
$$\sqrt{a^2 +b^2} \left(1-\frac{a b \cos (\theta ) \sin (\phi )}{a^2+b^2}-\frac{a^2 b^2 \cos ^2(\theta ) \sin ^2(\phi )}{2 \left(a^2+b^2\right)^2}-\frac{a^3 b^3 \cos ^3(\theta ) \sin ^3(\phi )}{2 \left(a^2+b^2\right)^3}-\cdots\right)$$
Now these functions can be integrated directly. (The odd powers of $\cos$ integrate to $0$ over the range here.)
For any $a,b$ you get some power series in $\left(\frac{a b}{a^2+b^2}\right)$
vis: $$\frac{\pi ^2 \sqrt{a^2+b^2}\left(2097152-\frac{262144 a^2 b^2}{\left(a^2+b^2\right)^2}-\frac{184320 a^4 b^4}{\left(a^2+b^2\right)^4}-\frac{268800 a^6 b^6}{\left(a^2+b^2\right)^6}-\frac{525525 a^8 b^8}{\left(a^2+b^2\right)^8}\right)}{1048576}$$
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