$\int_{0}^\pi\frac{\sin^2(\theta)}{(a-b\cos(\theta))^{3/2}}d\theta$

Question

As stated in the title. $$\int_{0}^\pi\frac{\sin^2(\theta)}{(a-b\cos(\theta))^{3/2}}d\theta$$ where $a,b$ are real, positive constants, $a \neq b$.

I tried Weierstrass substitution, $\theta=\frac{\pi}{2}-u$, trig. identities etc. but they all seem to yield a fourth order polynomial which seems analytically intractable. Is there a (easy?) way to evaluate the integral?

Answer 1

For the integrand to be real over $[0,\pi]$ we require $a>b$. Then using Byrd & Friedman 291.08 – the "most general form" for integrals containing $\sqrt{a-b\cos\theta}$ – we get after some simplification $$I=\frac8{(a+b)^{3/2}}\int_0^{K(m)}\operatorname{sn}^2u\operatorname{cd}^2u\,du$$ where the parameter $m=\frac{2b}{a+b}$. In turn, B&F 361.34 solves this as (again, after simplifying) $$I=\frac{2\sqrt{a+b}}{b^2}((2-m)K(m)-2E(m))$$