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.
 A: This calls for Kepler's angle! Let
$$\sin\theta=\frac{\sqrt{1-a^2}\sin\psi}{1+a\cos\psi}$$
Then
$$\begin{align}\cos\theta&=\frac{\cos\psi+a}{1+a\cos\psi}\\
d\theta&=\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\
1-a\cos\theta&=\frac{1-a^2}{1+a\cos\psi}\end{align}$$
Also when $\theta$ makes a full cycle, so does $\psi$, so
$$\begin{align}\int_0^{2\pi}\frac{\sin^2\theta}{\left(1-a\cos\theta\right)^3}d\theta&=\int_0^{2\pi}\frac{\left(1-a^2\right)\sin^2\psi}{\left(1+a\cos\psi\right)^2}\frac{\left(1+a\cos\psi\right)^3}{\left(1-a^2\right)^3}\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\
&=\frac1{\left(1-a^2\right)^{3/2}}\frac12\int_0^{2\pi}\left(1-\cos2\psi\right)d\psi\\
&=\frac1{2\left(1-a^2\right)^{3/2}}\left.\left[\psi-\frac12\sin2\psi\right]\right|_0^{2\pi}=\frac{\pi}{\left(1-a^2\right)^{3/2}}\end{align}$$
A: Note that
$$D = 2\int_0^{\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$$
Let $t=\tan\frac{\theta}2$, we have
\begin{align}
D&=2\int_0^{\infty} \frac{\frac{4t^2}{(1+t^2)^2}}{(1-a\frac{1-t^2}{1+t^2})^3}\frac{2}{1+t^2}\,dt\\
&=\int_0^{\infty} \frac{16t^2}{(1-a+(1+a)t^2)^3}\,dt\\
&=\frac{\pi}{(1-a^2)^{3/2}}\tag{1}
\end{align}
$(1):$ for all $a,b>0$,
\begin{align}
\int_0^{\infty} \frac{16t^2}{(a+bt^2)^3}\,dt&=\frac{1}{(ab)^{3/2}}\int_0^{\infty} \frac{16t^2}{(1+t^2)^3}\,dt\\
&=\frac{1}{(ab)^{3/2}}\int_0^{\infty} {8s^{1/2}}{(1+s)^{-3}}\,ds\tag{$s=t^2$}\\
&=\frac{\pi}{(ab)^{3/2}}\tag{2}
\end{align}
$(2):$ For $x,y>0$, Euler integral gives
$$\int^\infty_0 \frac {t^{x-1}}{(1+t)^{x+y }}dt =\frac {\Gamma (x)\Gamma (y)}{\Gamma (x+y)}$$
and for $x=y=\frac32$, we have
$$\frac {\Gamma^2 (\frac32)}{\Gamma (3)}=\frac\pi8$$
A: It is not difficult to check that for any $a\in\mathbb{R}$ such that $|a|<1$ we have
$$ \int_{0}^{2\pi}\frac{d\theta}{1-a\cos\theta} = \frac{2\pi}{\sqrt{1-a^2}}\tag{1}$$
hence by differentiating both sides of $(1)$ with respect to $a$ we have
$$ \int_{0}^{2\pi}\frac{\cos(\theta)\,d\theta}{(1-a\cos\theta)^2}=\frac{2a\pi}{(1-a^2)^{3/2}},\qquad \int_{0}^{2\pi}\frac{d\theta}{(1-a\cos\theta)^2}= \frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$
and by differentiating again
$$ \int_{0}^{2\pi}\frac{\sin^2(\theta)\,d\theta}{(1-a\cos\theta)^3}=\frac{\pi}{(1-a^2)^{3/2}} \tag{3}$$
is simple to prove.
A: Let $z=e^{i\theta}$ and then
$$ \cos\theta=\frac12(z+z^{-1}),\sin\theta=\frac1{2i}(z-z^{-1}),d\theta=\frac{1}{iz}dz. $$
So
\begin{eqnarray}
&&\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta\\
&=&\int_{|z|=1}\frac{[\frac1{2i}(z-z^{-1})]^2}{(1-\frac a2(z+z^{-1}))^3}\frac{1}{iz}dz\\
&=&-\int_{|z|=1}\frac{2 i \left(z^2-1\right)^2}{\left(a z^2+a-2 z\right)^3}dz\\
&=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z^2-\frac{2}{a} z+1\right)^3}dz\\
&=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3\left(z-\frac{1-\sqrt{1-a^2}}{a}\right)^3}dz\\
&=&-\frac{2i}{a^3}\cdot2\pi i\cdot\frac12\frac{d^2}{dz^2}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3}\bigg|_{z=\frac{1-\sqrt{1-a^2}}{a}}\\
&=&\frac{4\pi}{a^3}\cdot\frac{a^3}{4(1-a^2)^{3/2}}\\
&=&\frac{\pi}{(1-a^2)^{3/2}}.
\end{eqnarray}
A: 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. 
