Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$  ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$  .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved.  So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.

For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$

Written with StackEdit.

UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand).  I have also added an answer based on Quanto's but with the intermediate steps written out.

NOTE: This is a continuation (the easy bit!) of M.Strochyk's answer.
using $$z = \frac{1-\sqrt{1-a^{2}}} {a}
\mathrm{and}\,\mathrm{ defining}\, Q=\sqrt{1-a^2}$$
so that $$\left(
z - \frac{1+\sqrt{1-a^{2}}} {a}
\right) = \frac{-2Q}{a}$$
We can analyze the three terms of the Residue separately and merge them later
$$$$
TERM 1
$$
\frac
{3 z}{{\left(
z - \frac{1+Q} {a}
\right)}^{3}} 
= \frac
{3  \frac{1-Q} {a}}
{{\left(
\frac{-2Q}{a}
\right)}^{3}} 
= \frac
{3 a^2 \left(1-Q \right) (-2Q)^{2}}
{{(-2Q)}^{5}} 
$$TERM 2$$
- \frac{3 {\left(3 z^{2} + 1\right)}}{{\left(z - \frac{1+Q }{a}\right)}^{4}} 
=- \frac{3a^4 {\left(3 (\frac{1-Q}{a})^{2} + 1\right)  (-2Q) }}
 {{\left(-2Q\right)}^{5}} 
$$
TERM 3
$$
+ \frac{6 {\left((z)^{3} + z\right)}}{{\left(z - \frac{1+Q}{a}\right)}^{5}}
=+ \frac{6a^5 {\left((\frac{1-Q}{a})^{3} + (\frac{1-Q} {a})\right)}}
{{\left(-2Q\right)}^{5}}
$$
Summing together the three terms
$$
= \frac
{3 a^2 \left(4Q^2-4Q^3 \right) 
+6Qa^2 {\left( 3(1-2Q+Q^2)+a^2\right)  } 
+6a^2 {\left((1-Q)^3+a^2(1-Q) \right)}
}
{{(-2Q)}^{5}} 
$$
reducing to
$$
= \frac
{ \left(12a^2Q^2-12a^2Q^3 \right) 
+{\left( 18a^2Q-36a^2Q^2+18a^2Q^3+6a^4Q\right)  } 
+6a^2 {\left((1-3Q+3Q^2-Q^3)+a^2-Qa^2 \right)}
}
{{(-2Q)}^{5}} 
$$
and
$$
= \frac
{ \left(12a^2Q^2-12a^2Q^3 \right) 
+{\left( 18a^2Q-36a^2Q^2+18a^2Q^3+6a^4Q\right)  } 
+ \left(6a^2-18a^2Q+18a^2Q^2-6a^2Q^3+6a^4-6a^4Q \right)
}
{{(-2Q)}^{5}} 
$$
then
$$
= \frac
{
-6a^2Q^2
+ 6a^2+6a^4 
}
{{(-2Q)}^{5}} 
$$
So
$$
=\frac{6}{32} \frac
{(
a^2Q^2
- a^2-a^4 )
}
{Q^5}
$$
Multiplying by $-8\pi/a^3$ gives
$$
=-\frac{3\pi}{2a} \frac
{(
Q^2
- 1-a^2 )
}
{Q^5}
$$
but $Q^2$ = $1-a^2$ so
$$
=-\frac{3\pi}{2a} \frac
{(
-2a^2 )
}
{Q^5}
$$
giving
$$
= \frac
{3a\pi }
{(1-a^2)^{5/2}}
$$
which is the hypothesised result.
