<s>[INST] 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.
 [/INST] Lemma 1: If $|x| < 1$, then we know that, $\frac{1}{1 - x} = \sum_{n = 0}^{\infty}x^n$. Using this, we can further show that, $\frac{1}{(1 - x)^3} = \sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}x^n$.
Proof: The first sum is just the GP. The second sum can be obtained by simply counting the different ways to get $x^n$ in $(\sum x^n)(\sum x^n)(\sum x^n)$.

Lemma 2: $$\int_{0}^{2\pi}\cos^{2k}\theta d\theta = {2k \choose k}\frac{2}{4^k}\pi$$
Proof: From Reduction Formula, we can see that,
$$\int_{0}^{2\pi}\cos^{2k}\theta d\theta = \frac{2k - 1}{2k}\int_{0}^{2\pi}\cos^{2k - 2}\theta d\theta$$
Continuing in this fashion, gives us the result.

Lemma 3: $$\int_{0}^{2\pi}\cos^{2k+1}\theta d\theta = 0$$
Proof: This result also follows from Reduction Formula.

Lemma 4: $$\frac{3}{(1 - x)^{5/2}} = \sum_{n = 0}^{\infty} (2n + 3)(2n+1){2n \choose n}\left(\frac{x}{4}\right)^n$$
Proof: We can show that, when $|x| < 1$,
$$\frac{1}{\sqrt{1 - x}} = \sum_{n = 0}^{\infty} {2n \choose n}\left(\frac{x}{4}\right)^n$$
Differentiating once, we obtain,
$$\frac{1}{(1 - x)^{3/2}} = \sum_{n = 1}^{\infty} n{2n \choose n}\left(\frac{x^{n - 1}}{4^n}\right) = \sum_{n = 0}^{\infty} (n + 1){2(n+1) \choose (n+1)}\left(\frac{x^{n}}{4^{n+1}}\right)$$
$$= \frac{1}{2}\sum_{n = 0}^{\infty} (2n + 1){2n \choose n}\left(\frac{x}{4}\right)^n$$
Differentiating again, we get,
$$\frac{3}{(1 - x)^{5/2}} = 2\sum_{n = 1}^{\infty} n(2n + 1){2n \choose n}\left(\frac{x^{n-1}}{4^{n}}\right) = \sum_{n = 0}^{\infty} (2n + 3)(2n+1){2n \choose n}\left(\frac{x}{4}\right)^n$$

Theorem: $$\int_{0}^{2\pi}\frac{\cos\theta}{(1 - a\cos\theta)^3}d\theta = \frac{3a\pi}{(1 - a^2)^{5/2}}$$
Proof: Given, $0 < a < 1$, therefore $|a \cos\theta| < 1$
 Hence, from Lemma 1, by replacing $x$ with $a\cos\theta$, we get
$$\frac{1}{(1 - a\cos\theta)^3} = \sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\cos^n\theta$$
This gives us,
$$\int_{0}^{2\pi}\frac{\cos\theta}{(1 - a\cos\theta)^3}d\theta = \int_{0}^{2\pi} \cos\theta\left(\sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\cos^n\theta\right)d\theta$$
$$=\sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\int_{0}^{2\pi}\cos^{n+1}\theta d\theta$$
Using Lemma 2 and 3 and setting $n = 2k + 1$, we get,
$$\sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}a^n\int_{0}^{2\pi}\cos^{n+1}\theta d\theta = \sum_{k = 0}^{\infty}\frac{(2k + 2)(2k + 3)}{2}a^{2k + 1} {2(k+1) \choose k+1}\frac{2}{4^{k+1}}\pi$$
$$=\frac{a\pi}{2}\sum_{k = 0}^{\infty}(k + 1)(2k + 3)a^{2k} {2(k+1) \choose k+1}\frac{1}{4^{k}}$$
$$= \frac{a\pi}{2}\sum_{k = 0}^{\infty}(2k+3)(2k+2)(2k+1)\frac{(2k)!}{k!(k+1)!}\left(\frac{a}{2}\right)^{2k}$$
$$= a\pi\sum_{k = 0}^{\infty}(2k+3)(2k+1)\frac{(2k)!}{k!k!}\left(\frac{a^2}{4}\right)^{k} = a\pi\sum_{k = 0}^{\infty}(2k+3)(2k+1){2k \choose k} \left(\frac{a^2}{4}\right)^{k}$$
Finally, we use Lemma 4 to obtain the result.
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