<s>[INST] Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral  for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[ 
\frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2}
+\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}  \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi}
.$$
which can be expressed as
$$D= \left[ 
\frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}  \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2}
\right]_0^{2\pi}
.$$
This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.
 [/INST] We can use the standard formula $$\int_{0}^{2\pi}\frac{dx}{1-a\cos x} =\frac{2\pi}{\sqrt{1-a^2}},|a|<1\tag{1}$$ Note that $$\frac{d} {dx} \frac{\sin x} {1-a\cos x} =\frac{\cos x-a} {(1-a\cos x) ^2} =-\frac{1}{a}\frac{a^2-1+1-a\cos x} {(1-a\cos x) ^2}$$ and integrating the above with respect to $x$ in interval $[0,2\pi]$ we get $$0=\frac{1-a^2}{a}\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}-\frac{1}{a}\int_{0}^{2\pi}\frac{dx}{1-a\cos x} $$ ie $$\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}=\frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$ Next we can observe that $$\frac{d} {dx} \frac{\sin x} {(1-a\cos x) ^2}=\frac{\cos x-a-a\sin^2x}{(1-a\cos x) ^3}$$ and integrating the above we obtain $$0=-aI+\frac{1-a^2}{a}\cdot J-\frac{1}{a}\cdot\frac{2\pi}{(1-a^2)^{3/2}}$$ ie $$(1-a^2)J-a^{2}I=\frac{2\pi}{(1-a^2)^{3/2}}\tag{3}$$ where $I$ is the integral in question and $J=\int_{0}^{2\pi}(1-a\cos x) ^{-3}\,dx$. We need another relationship between $I, J$ to evaluate both of them. Note that if $t=\cos x$ then $$\frac{\sin^2x}{(1-a\cos x) ^3}=\frac{1-t^2}{(1-at)^3}=\frac{A} {1-at}+\frac{B}{(1-at)^{2}}+\frac{C}{(1-at)^3}\tag{4}$$ where $$A=-\frac{1}{a^2},B=\frac{2}{a^2},C=-\frac{1-a^2}{a^2}$$ and on integrating equation $(4)$ we get $$a^2I+(1-a^2)J=\frac{2\pi(1+a^2)}{(1-a^2)^{3/2}}\tag{5}$$ Subtracting equation $(3)$ from equation $(5)$ we get $$I=\frac{\pi} {(1-a^2)^{3/2}}$$ Comparing this elementary solution with the one offered by Jack D'Aurizio shows us the power of Feynman's technique.

Formula $(1)$ is an easy consequence of the following result $$\int_{0}^{\pi}\frac{dx}{a-b\cos x} =\frac{\pi} {\sqrt{a^2-b^2}},a>|b|\tag{6}$$ which is (not so) easily proved via the substitution $$(a-b\cos x) (a+b\cos y) =a^2-b^2$$ The same substitution $$(1-a\cos x) (1+a\cos y) =1-a^2$$ has been used directly in this answer to evaluate the integral in question with much less effort. 
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