# 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 $$0). $$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 $$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta= \frac{-\sin\theta}{(1-a\cos\theta)^3} +\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.$$

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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.

• I haven't tried it because I'm on my phone, but this looks like an integral that you could rewrite as a complex integral over the unit circle and evaluate with the residue theorem. Here $z=\cos(\theta)+i\sin(\theta)$, and note that $\cos(\theta)=(z+\overline z)/2=(z+1/z)/2$.
– a...
Commented Oct 25, 2014 at 20:28
• Thanks for the suggestion. M.Strochyk's answer develops on this. Commented Oct 27, 2014 at 12:32

According to Brian Bóruma's suggestion, integral can be evaluated using the residue theorem $$\int\limits_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,{d\theta}=\dfrac{1}{2}\int\limits_{|z|=1}{\dfrac{z+\frac{1}{z}}{\left[1-\frac{a}{2}\left(z+\frac{1}{z}\right)\right]^3} \frac{dz}{iz}}=\\ =-4i\int\limits_{|z|=1}{\dfrac{z^3+z}{\left(-az^2+2z-a\right)^3}dz}= \\ =-4i\int\limits_{|z|=1}{\dfrac{z^3+z}{-a^3\left(z-z_1\right)^3 \left(z-z_2\right)^3}dz}=-\dfrac{8\pi}{a^3}{\operatorname{Res}\limits_{z=z_1}{\dfrac{z^3+z}{\left(z-z_1\right)^3 \left(z-z_2\right)^3}}},$$ where $z_1=\dfrac{1-\sqrt{1-a^2}}{a}, \;\; z_2=\dfrac{1+\sqrt{1-a^2}}{a}.$ Only the point $z_1$ lies in the unit disc and it is a third-order pole for the integrand function.
The residue can be calculated in a standard way: $$\operatorname{Res}\limits_{z=z_1}{\dfrac{z^3+z}{\left(z-z_1\right)^3 \left(z-z_2\right)^3}}=\\= \dfrac{1}{2!} \lim\limits_{z\to z_1}{\dfrac{d^2}{dz^2}\left(\dfrac{\left(z-z_1\right)^3 \left(z^3+z \right)}{\left(z-z_1\right)^3 \left(z-z_2\right)^3} \right)} =\dfrac{1}{2!} \lim\limits_{z\to z_1}{\dfrac{d^2}{dz^2}\left(\dfrac{z^3+z }{ \left(z-z_2\right)^3} \right)} = \\ =\left. \left[\frac{3 z}{{\left(z - \frac{1+\sqrt{1-a^{2}}} {a}\right)}^{3}} - \frac{3 {\left(3 z^{2} + 1\right)}}{{\left(z - \frac{1+\sqrt{1-a^{2}} }{a}\right)}^{4}} + \frac{6 {\left(z^{3} + z\right)}}{{\left(z - \frac{1+\sqrt{1-a^{2}}}{a}\right)}^{5}}\right] \right|_{z=z_1}.$$

• I edited my answer. Commented Oct 26, 2014 at 22:50
• Magnificent! (Alhough I personally don't comprehend the residue method). Commented Oct 27, 2014 at 11:41
• I have added a continuation to your analysis as a separate answer for now. If you would like me to incorporate it in your answer please let me know. Commented Oct 27, 2014 at 11:43
• I think it is better to place your work into a separate answer. Commented Oct 27, 2014 at 21:03

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.

• Why set $n-1 =2k$? Shouldn't it be $n+1=2k$, to match "cos^{n+1}$in previous line? Commented Oct 26, 2014 at 8:50 • Either is fine, if you set n+1 = 2k, then in subsequent steps, you simply use n = 2k - 1. Commented Oct 26, 2014 at 8:53 • Looking good :) But sorry I don't understand how you replace$\cos^{n+1}$by$\cos^{2k}$which equals$\cos^{n-1}$. And please can you explain:- (a) how does the first$\Sigma$expression arrive? (b) how does the binomial expression arrive? Commented Oct 26, 2014 at 21:37 • I didn't replace$\cos^{n+1}$with$\cos^{2k}$. I replaced it with$\cos^{2k+2}. I just wrote the formula for general even powers. For (a) and (b), I have updated the answer with details on how to prove them, they are quite simple to prove. Commented Oct 27, 2014 at 18:49 • Many thanks for the details. Commented Oct 27, 2014 at 22:03 With $$\int_{0}^{2\pi} \frac{1}{1-a \cos \theta} d \theta=\frac{2\pi}{\sqrt{1-a^2}}$$ evaluate the following integrals successively \begin{align}\int_{0}^{2\pi} \frac{1}{(1-a \cos \theta)^2} d\theta =&\frac{d}{da} \int_{0}^{2\pi} \frac{a}{1-a \cos \theta} d \theta=\frac{2\pi}{(1-a^2)^{3/2}}\\ \int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^3} d\theta =&\frac{d}{da} \int_{0}^{2\pi} \frac{1}{2(1-a \cos \theta)^2 }d \theta=\frac{3a\pi}{(1-a^2)^{5/2}}\\ \int_0^{2\pi} \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta =&\int_0^{2\pi} \sin\theta \ d\left(-\frac{1}{(1-a\cos \theta)^3}\right)\\ \overset{ibp}=&\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}d\theta=\frac{3a\pi}{(1-a^2)^{5/2}}\\ \end{align} • @steveOw $$\frac{d}{da} \frac{2\pi a}{\sqrt{1-a^2}}=\frac{2\pi}{(1-a^2)^{3/2}}$$ Commented Jul 11, 2022 at 20:12 • Also, I dont know why line 4 (3 a \sin^2(\theta)$...) is needed or where the RHS comes from. Commented Jul 11, 2022 at 22:47 • @steveOw - corrected line 3 Commented Jul 12, 2022 at 11:40 • For the Line 3 correction, thanks. But my confusion (see my previous comment) about Line 4 remains. Commented Jul 12, 2022 at 14:55 • @steveOw - Line 4 is the same answer I provided to your question in the linked post, which is similar to yours. Commented Jul 12, 2022 at 21:27 I try to tackle the second integral $$\int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta$$ by the result in my post $$I_n(a)=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{(-1)^{n-1} \pi}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{1}{\sqrt{a^{2}-1}}\right)$$ I start with the integral $$\int_{0}^{\pi} \frac{1}{(1-a \cos \theta)^{2}} d \theta =\frac{1}{a^{2}} \int_{0}^{\pi} \frac{1}{\left(\frac{1}{a}-\cos \theta\right)^{2}} d \theta=\frac{1}{a^{2}}\left(\frac{\pi\left(\frac{1}{a}\right)}{\left(\frac{1}{a^{2}}-1\right)^{\frac{3}{2}}}\right) =\frac{\pi}{\left(1-a^{2}\right)^{\frac{3}{2}}}$$ Differentiating both sides w.r.t. $$a$$ yields $$2 \int_{0}^{\pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta =\frac{3 \pi a}{\left(1-a^{2}\right)^{\frac{5}{2}}}$$ By symmetry, we get $$\boxed{\int_{0}^{2 \pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta=2 \int_{0}^{\pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta= \frac{3 \pi a}{\left(1-a^{2}\right)^{\frac{5}{2}}}}$$ • Taking your formula from the linked post as a given, for me this is a nice simple proof. Thankyou. Commented Jul 11, 2022 at 22:53 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.

Here I have taken the compact answer by /u/Quanto (to whom all credit is due) and simply filled in some intermediate steps. His trick was to (twice) differentiate inside the integral ( a special case of the Leibniz integral rule ).

Using the known definite integral:- $$\int_0^{2\pi}\frac{1}{E+F\cos x}\mathrm{d}x= \frac{2\pi}{\sqrt{E^2-F^2}}$$ with $$E=1$$, $$F=-a$$ and $$x=\theta$$, we can write:-

$$\int_{0}^{2\pi} \frac{1}{1-a \cos \theta} d \theta=\frac{2\pi}{\sqrt{1-a^2}}$$ We can multiply both sides by $$a$$ (so long as $$a$$ is not a function of $$\theta$$) to give:- $$\int_{0}^{2\pi} \frac{a}{1-a \cos \theta} d \theta=\frac{2a\pi}{\sqrt{1-a^2}}$$ now differentiate both sides (within the integral on the LHS) with respect to $$a$$:-

$$\int_{0}^{2\pi} \frac{1}{(1-a \cos \theta)^2} d\theta =\frac{2\pi}{(1-a^2)^{3/2}}$$

again, differentiate both sides (within the integral on the LHS) with respect to $$a$$:-

$$\int_{0}^{2\pi} \frac{2\cos \theta}{(1-a \cos \theta)^3} d\theta =\frac{6a\pi}{(1-a^2)^{5/2}}$$

Hence:- $$\int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^3} d\theta =\frac{3a\pi}{(1-a^2)^{5/2}}.$$