<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] 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
$$\frac {\Gamma^2 (\frac32)}{\Gamma (3)}=\frac\pi8$$
</s>