Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ 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 when I test (using www.WolframAlpha.com) for 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
$$
D_2
\, =\,
\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$

How could I go about proving:-
(1) $D_1$ = $D_2$, 
(SOLVED, I think, by my two answers below, but using WolframAlpha to obtain integral solutions)
$$$$
(2) $D_1$ = $R$ or $D_2$ = $R$. 
(MOVED to a separate question:
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$).

UPDATE 1
You can see how WolframAlpha
produces these results by inputting the following input texts:-
For Eqtn 1 with a=0.1 input: integrate (3*0.1(sinx)^2)/((1-0.1*cosx)^4) from x=0 to 2*pi
For Eqtn 2 with a=0.1 input: integrate (cosx)/((1-0.1*cosx)^3) from x=0 to 2*pi
for Result with a=0.1 input: evaluate 3 0.1 pi/(1-0.1^2)^(5/2)

UPDATE 2
WolframAlpha also computes expressions for the indefinite integrals as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
$$$$
$$$$
$$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}}
$$
Note that the final terms of each expression are equivalent to each other.  This could be useful. For example we can define a difference function $f_3 = f_1-f_2$ whose indefinite integral $I_3 = I_1-I_2$ will exclude the common awkward third term.  
Let us assume that $f_3$ is continuously integrable over the range $0,2\pi$ (we cannot be sure by inspection alone, but it can be shown, see my answer below).  Then, if $D_1=D_2$ over the range $0,2\pi$ then $D_1-D_2=0$ and so $D_3$ (=$\int_0^{2\pi}f_3\,d\theta$) should have value zero.  This is expanded on in my answer below.
 A: Hint:
Observe that,
$$\frac{\partial}{\partial a}\left[\frac{1}{\left(1-a\cos{\theta}\right)^2}\right]=\frac{2\cos{\theta}}{\left(1-a\cos{\theta}\right)^3}.$$
Thus, we can simplify the integral we have to compute via the technique of differentiating under the integral sign:
$$\begin{align}
I(a)
&=\int_{0}^{2\pi}\frac{\cos{\theta}}{\left(1-a\cos{\theta}\right)^3}\,\mathrm{d}\theta\\
&=\frac12\int_{0}^{2\pi}\frac{\partial}{\partial a}\left[\frac{1}{\left(1-a\cos{\theta}\right)^2}\right]\,\mathrm{d}\theta\\
&=\frac12\frac{\partial}{\partial a}\int_{0}^{2\pi}\frac{\mathrm{d}\theta}{\left(1-a\cos{\theta}\right)^2}.\\
\end{align}$$
A: (as per David H's suggestion to use Integration By Parts)
Hypothesis: The two definite integrals $D_1$ and $D_2$ have the same value, i.e. $D_1 = D_2$.
Using integration by parts
$$
\int uv' \,d\theta= uv-\int u'v \,d\theta
$$
let us define 
$$
u = \frac{1}{(1-a\cos\theta)^3}
\qquad
v' = \cos\theta
$$
So ($u'$ obtained from WolframAlpha)
$$
u'=\frac{-3a\sin\theta}{(1-a\cos\theta)^4}
\qquad
v=\sin\theta
$$
Then 
$$
\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta= uv - \int u'v\, d\theta
$$
$$=\frac{-\sin\theta}{(1-a\cos\theta)^3}
-\int \frac{-3a\sin^2\theta}{(1-a\cos\theta)^4}
$$
So
$$\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.
$$
Now consider the definite integrals over the range $0,2\pi$
$$\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\left[\frac{-\sin\theta}{(1-a\cos\theta)^3}\right]_0^{2\pi}
+\int_0^{2\pi} \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta
$$
The term in square brackets clearly goes to zero over the range $0,2\pi$ because $\sin(0)=\sin(2\pi)=0$ and therefore we have $D_1=D_2$.
