Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$

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I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$: $$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$

which gives (Wolfram Alpha)

$$D= \left[ \frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2} +\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi} .$$

which can be expressed as $$D= \left[ \frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2} \right]_0^{2\pi} .$$

This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.

UPDATE

I accepted Jack's answer although I did not follow all the steps. In particular the first equation:-

It is not difficult to check that for any $a\in\mathbb{R}$ such that $|a|<1$ we have $$ \int_{0}^{2\pi}\frac{d\theta}{1-a\cos\theta} = \frac{2\pi}{\sqrt{1-a^2}}\tag{1}$$

but I found an explanation for this in the Kepler-based answer from \u\user5713492 to this question re solving $\int\frac{dx}{a+b\cos{x}}$. Other responders also referred to this equation.

Also interetsing is this article - warning PDF! linked to by \u\Adhvaitha.

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It is not difficult to check that for any $a\in\mathbb{R}$ such that $|a|<1$ we have $$ \int_{0}^{2\pi}\frac{d\theta}{1-a\cos\theta} = \frac{2\pi}{\sqrt{1-a^2}}\tag{1}$$ hence by differentiating both sides of $(1)$ with respect to $a$ we have $$ \int_{0}^{2\pi}\frac{\cos(\theta)\,d\theta}{(1-a\cos\theta)^2}=\frac{2a\pi}{(1-a^2)^{3/2}},\qquad \int_{0}^{2\pi}\frac{d\theta}{(1-a\cos\theta)^2}= \frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$ and by differentiating again $$ \int_{0}^{2\pi}\frac{\sin^2(\theta)\,d\theta}{(1-a\cos\theta)^3}=\frac{\pi}{(1-a^2)^{3/2}} \tag{3}$$ is simple to prove.

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Let $z=e^{i\theta}$ and then $$ \cos\theta=\frac12(z+z^{-1}),\sin\theta=\frac1{2i}(z-z^{-1}),d\theta=\frac{1}{iz}dz. $$ So \begin{eqnarray} &&\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta\\ &=&\int_{|z|=1}\frac{[\frac1{2i}(z-z^{-1})]^2}{(1-\frac a2(z+z^{-1}))^3}\frac{1}{iz}dz\\ &=&-\int_{|z|=1}\frac{2 i \left(z^2-1\right)^2}{\left(a z^2+a-2 z\right)^3}dz\\ &=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z^2-\frac{2}{a} z+1\right)^3}dz\\ &=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3\left(z-\frac{1-\sqrt{1-a^2}}{a}\right)^3}dz\\ &=&-\frac{2i}{a^3}\cdot2\pi i\cdot\frac12\frac{d^2}{dz^2}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3}\bigg|_{z=\frac{1-\sqrt{1-a^2}}{a}}\\ &=&\frac{4\pi}{a^3}\cdot\frac{a^3}{4(1-a^2)^{3/2}}\\ &=&\frac{\pi}{(1-a^2)^{3/2}}. \end{eqnarray}

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Note that $$D = 2\int_0^{\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$$ Let $t=\tan\frac{\theta}2$, we have \begin{align} D&=2\int_0^{\infty} \frac{\frac{4t^2}{(1+t^2)^2}}{(1-a\frac{1-t^2}{1+t^2})^3}\frac{2}{1+t^2}\,dt\\ &=\int_0^{\infty} \frac{16t^2}{(1-a+(1+a)t^2)^3}\,dt\\ &=\frac{\pi}{(1-a^2)^{3/2}}\tag{1} \end{align} $(1):$ for all $a,b>0$, \begin{align} \int_0^{\infty} \frac{16t^2}{(a+bt^2)^3}\,dt&=\frac{1}{(ab)^{3/2}}\int_0^{\infty} \frac{16t^2}{(1+t^2)^3}\,dt\\ &=\frac{1}{(ab)^{3/2}}\int_0^{\infty} {8s^{1/2}}{(1+s)^{-3}}\,ds\tag{$s=t^2$}\\ &=\frac{\pi}{(ab)^{3/2}}\tag{2} \end{align} $(2):$ For $x,y>0$, Euler integral gives $$\int^\infty_0 \frac {t^{x-1}}{(1+t)^{x+y }}dt =\frac {\Gamma (x)\Gamma (y)}{\Gamma (x+y)}$$ and for $x=y=\frac32$, we have $$\frac {\Gamma^2 (\frac32)}{\Gamma (3)}=\frac\pi8$$

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This calls for Kepler's angle! Let $$\sin\theta=\frac{\sqrt{1-a^2}\sin\psi}{1+a\cos\psi}$$ Then $$\begin{align}\cos\theta&=\frac{\cos\psi+a}{1+a\cos\psi}\\ d\theta&=\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\ 1-a\cos\theta&=\frac{1-a^2}{1+a\cos\psi}\end{align}$$ Also when $\theta$ makes a full cycle, so does $\psi$, so $$\begin{align}\int_0^{2\pi}\frac{\sin^2\theta}{\left(1-a\cos\theta\right)^3}d\theta&=\int_0^{2\pi}\frac{\left(1-a^2\right)\sin^2\psi}{\left(1+a\cos\psi\right)^2}\frac{\left(1+a\cos\psi\right)^3}{\left(1-a^2\right)^3}\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\ &=\frac1{\left(1-a^2\right)^{3/2}}\frac12\int_0^{2\pi}\left(1-\cos2\psi\right)d\psi\\ &=\frac1{2\left(1-a^2\right)^{3/2}}\left.\left[\psi-\frac12\sin2\psi\right]\right|_0^{2\pi}=\frac{\pi}{\left(1-a^2\right)^{3/2}}\end{align}$$

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We can use the standard formula $$\int_{0}^{2\pi}\frac{dx}{1-a\cos x} =\frac{2\pi}{\sqrt{1-a^2}},|a|<1\tag{1}$$ Note that $$\frac{d} {dx} \frac{\sin x} {1-a\cos x} =\frac{\cos x-a} {(1-a\cos x) ^2} =-\frac{1}{a}\frac{a^2-1+1-a\cos x} {(1-a\cos x) ^2}$$ and integrating the above with respect to $x$ in interval $[0,2\pi]$ we get $$0=\frac{1-a^2}{a}\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}-\frac{1}{a}\int_{0}^{2\pi}\frac{dx}{1-a\cos x} $$ ie $$\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}=\frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$ Next we can observe that $$\frac{d} {dx} \frac{\sin x} {(1-a\cos x) ^2}=\frac{\cos x-a-a\sin^2x}{(1-a\cos x) ^3}$$ and integrating the above we obtain $$0=-aI+\frac{1-a^2}{a}\cdot J-\frac{1}{a}\cdot\frac{2\pi}{(1-a^2)^{3/2}}$$ ie $$(1-a^2)J-a^{2}I=\frac{2\pi}{(1-a^2)^{3/2}}\tag{3}$$ where $I$ is the integral in question and $J=\int_{0}^{2\pi}(1-a\cos x) ^{-3}\,dx$. We need another relationship between $I, J$ to evaluate both of them. Note that if $t=\cos x$ then $$\frac{\sin^2x}{(1-a\cos x) ^3}=\frac{1-t^2}{(1-at)^3}=\frac{A} {1-at}+\frac{B}{(1-at)^{2}}+\frac{C}{(1-at)^3}\tag{4}$$ where $$A=-\frac{1}{a^2},B=\frac{2}{a^2},C=-\frac{1-a^2}{a^2}$$ and on integrating equation $(4)$ we get $$a^2I+(1-a^2)J=\frac{2\pi(1+a^2)}{(1-a^2)^{3/2}}\tag{5}$$ Subtracting equation $(3)$ from equation $(5)$ we get $$I=\frac{\pi} {(1-a^2)^{3/2}}$$ Comparing this elementary solution with the one offered by Jack D'Aurizio shows us the power of Feynman's technique.


Formula $(1)$ is an easy consequence of the following result $$\int_{0}^{\pi}\frac{dx}{a-b\cos x} =\frac{\pi} {\sqrt{a^2-b^2}},a>|b|\tag{6}$$ which is (not so) easily proved via the substitution $$(a-b\cos x) (a+b\cos y) =a^2-b^2$$ The same substitution $$(1-a\cos x) (1+a\cos y) =1-a^2$$ has been used directly in this answer to evaluate the integral in question with much less effort.

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By integration by parts, we have $$ \begin{aligned}\int_0^{2 \pi} \frac{\sin ^2 x}{(1-a \cos x)^3} d x =- & \frac{1}{2 a} \int_0^{2 \pi} \sin x d\left[\frac{1}{(1-a \cos x)^2}\right] \\ = & {\left[-\frac{\sin x}{2 a(1-a \cos x)^2}\right]_0^{2 \pi}+\frac{1}{2 a} \int_0^{2 \pi} \frac{\cos x}{(1-a \cos x)^2} d x } \\ = & \frac{1}{2 a} \frac{d}{d a}\left[\int_0^{2 \pi} \frac{1}{1-a \cos x} d x\right] \\ = & \frac{1}{2 a} \cdot \frac{d}{d a}\left(\frac{2 \pi}{\sqrt{1-a^2}}\right) \\ = & \frac{1}{2 a} \cdot \frac{2 a \pi}{\left(1-a^2\right)^{\frac{3}{2}}} \\ = & \frac{\pi}{\left(1-a^2\right)^{\frac{3}{2}}} \end{aligned} $$