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 $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$).


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)


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.

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    $\begingroup$ Have you taken any Complex Analysis class? $\endgroup$ Commented Oct 20, 2014 at 20:04
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    $\begingroup$ $${\sin^{2}\left(\theta\right) \over \left[1 - \cos\left(\theta\right)\right]^{4}} \sim {16 \over \theta^{6}}\ \mbox{when}\ \theta\sim 0$$. Do they converge ?. $\endgroup$ Commented Oct 20, 2014 at 20:22
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    $\begingroup$ Can you include some of your WA testing? As Felix has just indicated, there are problems of convergence with these integrals... $\endgroup$ Commented Oct 20, 2014 at 20:24
  • $\begingroup$ Apologies, I missed a factor "a" from the LHS denominators. Now corrected. $\endgroup$
    – steveOw
    Commented Oct 20, 2014 at 21:41
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    $\begingroup$ You can at least prove that the two integrals are equal by a simple integration by parts, though this doesn't tell you what the value is. $\endgroup$
    – David H
    Commented Oct 20, 2014 at 21:46

3 Answers 3



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:

$$\begin{align} I(a) &=\int_{0}^{2\pi}\frac{\cos{\theta}}{\left(1-a\cos{\theta}\right)^3}\,\mathrm{d}\theta\\ &=\frac12\int_{0}^{2\pi}\frac{\partial}{\partial a}\left[\frac{1}{\left(1-a\cos{\theta}\right)^2}\right]\,\mathrm{d}\theta\\ &=\frac12\frac{\partial}{\partial a}\int_{0}^{2\pi}\frac{\mathrm{d}\theta}{\left(1-a\cos{\theta}\right)^2}.\\ \end{align}$$

  • $\begingroup$ Thanks. I am reading up on "differentiation under the integral" and hopefully will apply it in my separate answer. But maybe instead use d/da[1/(1 - a cos \theta)^3]? $\endgroup$
    – steveOw
    Commented Oct 21, 2014 at 11:36
  • $\begingroup$ Differentiation under the integral is beyond my present ability so I have "cheated" and used WolframAlpha (see my answer). I would be interested to see how d-u-t-i can be applied to this problem but it looks like it will be a lot of effort. Cheers :-). – $\endgroup$
    – steveOw
    Commented Oct 23, 2014 at 10:10
  • $\begingroup$ Aha, I was thinking this hint referred to integrating by parts (your earlier comment) but now I see IBP solves the first part of the problem, quite simply. I will do another answer based on IBP (unless you would like to do it yourself?). $\endgroup$
    – steveOw
    Commented Oct 24, 2014 at 18:12


Consider whether the Difference Function $\,f_3=f_1-f_2$ integrates to zero over the interval $0,2\pi$. If it does then $D_1 = D_2$.


The difference function does integrate to zero over the given interval. Therefore the two definite integrals $D_1$ and $D_2$ are equal.


Let us try to prove the hypothesis that the two definite integral expressions $D_1$, $D_2$ are equal for all values of $a$ (where $0<a<1$), i.e. $$ D_1= \int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4} d\theta =\int_0^{2\pi}\frac{(1-a \cos\theta)\cos \theta}{(1-a\cos \theta)^4}d\theta =D_2 $$

Let us define functions $f_1$, $f_2$ and their respective indefinite integrals $I_1$ and $I_2$ such that $$ I_1(a,\theta) = \int f_1(a,\theta) \, \mathrm{d}\theta =\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4} d\theta $$ and $$I_2(a,\theta) = \int f_2(a,\theta) \, \mathrm{d}\theta =\int\frac{(1-a \cos\theta)\cos \theta}{(1-a\cos \theta)^4}d\theta $$

WolframAlpha gives us for the indefinite integrals $I_1$ and $I_2$ $$I_1(a,\theta) \, =\, \int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta \,=\, $$ $$constant1 + \frac {a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]} {2(a^2-1)^{5/2}(a\cos\theta-1)^3} $$

$$-\frac {6a\,(a\cos\theta-1)^3\,\tanh^-1 \left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right) } {(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3} $$ and $$I_2(a,\theta) \, =\, \int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta \, =\, $$ $$constant2 - \frac {2a^2\sin\theta-sin\theta} {2(a^2-1)^2(a\cos\theta-1)} -\frac {\sin\theta} {2(a^2-1)(a\cos\theta-1)^2} $$

$$ -\frac {3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)} {(a^2-1)^{5/2}} $$

If the functions $I_1$ and $I_2$ were continuous over the interval $0,2\pi$ we could simply test whether the following statement is true: $$I_1(a,2\pi) - I_1(a,0) = I_2(a,2\pi) - I_2(a,0)$$ However the functions $I_1$ and $I_2$ are not continuous over the interval $0,2\pi$. This is due to the incorporation, in both of these functions, of the awkward term involving $\tanh^1()$ and $\tan(\theta/2)$ which produces a singularity when $\theta/2 = \pi/2$.

As an alternative approach let us obtain the function $I_3$ which is the indefinite integral of the Difference Function $f_3$ defined by $$f3(a,\theta) =f1(a,\theta)-f2(a,\theta)$$

As mentioned in the Question ( Update 2 ) subtraction of $I_2$ from $I_1$ will eliminate from $I_3$ the awkward final term in $\tanh^-1()$ which is common to both $I_1$ and $I_2$. Presumably then, although I haven't proved it, we could find a simple expression for $I_3$ by algebraic rationalisation of $I_1 - I_2$.

Conveniently however, when given the formula for $I_3$, WolframAlpha returns a simple result thus

$$I_3(a,\theta) = \int\frac{3a\sin^2\theta - \cos \theta(1-a \cos\theta)} {(1-a\cos \theta)^4} d\theta =\frac{sin\theta}{(a\cos\theta-1)^3}+ constant3 $$

Now, to prove that $D_1=D_2$ we simply need to show that $D_3=0$. This can be done by showing that $I_3(a,2\pi)-I_3(a,0) = 0$.

Knowing that $\sin(2\pi)=\sin(0)=0$ and that $\cos(2\pi)=\cos(0)=1$, it is simple to show that

$$ I_3(a,2\pi)=\frac{sin(2\pi)}{(a\cos(2\pi)-1)^3}+ constant3= 0+ constant3 $$ and $$ I_3(a,0)=\frac{sin(0)}{(a\cos(0)-1)^3}+ constant3= 0+ constant3 $$ So $$ I_3(a,2\pi)-I_3(a,0) = 0+ constant3 -(0+ constant3) = 0 $$ Therefore the original hypothesis is confirmed: $$ \int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4} d\theta =\int_0^{2\pi}\frac{(1-a \cos\theta)\cos \theta}{(1-a\cos \theta)^4}d\theta $$


Note: Part (2) of the question is not answered by this.


(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 $D_1=D_2$.


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