Dtyjlui

I'm having a hard time trying to resolve this integral :
$$int_{0}^{2pi}frac{mathrm dx}{(1-acos x)^2}$$
where $a$ is a positive real constant.

I tried using substitution, but I'm stuck by the fact that the integral must be computed between $0$ and $2pi$, which leads to integration between $X$ and $X$ (where $X$ can be $0$ or $1$ anything following the substitution in the boundaries).
I'm not looking for the full result if the computation is complicated, just some lead to start off...

Thanks

UPDATE : seeing the answers proposed, I checked that indeed $|a|<1$.

The polar equation of an ellipse (with respect to a focus, taking the aphelion as the point associated to $theta=0$) is
$$ rho(theta) = frac{ell}{1-ecostheta} $$
where, in terms of the standard parameters, $ell$ is the semi-latus rectum $frac{b^2}{a}$ and $e$ is the eccentricity $frac{c}{a}$.

The area enclosed by such ellipse is $pi a b $ or
$$ frac{1}{2}int_{0}^{2pi}frac{ell^2}{(1-ecostheta)^2},dtheta, $$
so we get that for any $ein[0,1)$ the following identity holds:
$$ int_{0}^{2pi}frac{dtheta}{(1-ecostheta)^2}= frac{2pi a b}{ell^2}=2pileft(frac{a}{b}right)^3=frac{2pi}{left(frac{b^2}{a^2}right)^{3/2}}={frac{2pi}{(1-e^2)^{3/2}}}. $$
The same holds if, in the LHS, we replace $e$ with $-e$, since $ rho(theta) = frac{ell}{1+ecostheta} $ is the polar equation of the same ellipse, with the perihelion being taken as the point associated to $theta=0$. So we get that for any $ain(-1,1)$
$$ int_{0}^{2pi}frac{dtheta}{(1-acostheta)^2}= color{blue}{frac{2pi}{(1-a^2)^{3/2}}}. $$

HINT:

Note that for $|a|<1$

$$begin{align}
int_0^{2pi}frac{1}{(1-acos(theta))^2},dtheta&=frac1{a^2}int_0^{2pi}frac{1}{(1/a-cos(theta))^2},dtheta\\
&=-frac1{a^2}frac{d}{d(1/a)}int_0^{2pi}frac{1}{1/a-cos(theta)},dthetatag1\\
&= -frac2{a^2}frac{d}{d(1/a)}int_0^{pi}frac{1}{1/a-cos(theta)},dthetatag2
end{align}$$

Now use the Weierstrass substitution to evaluate the integral on the right-hand side of $(2)$ and differentiate to obtain the coveted result.

Alternatively, one could use contour integration to evaluate the integral of interest or evaluate the integral on the right-hand side of $(1)$.

My goal with this answer is to give you a bunch of different really general integral formulas that you may find very useful.

$$F(a)=int_0^{pi}frac{mathrm dx}{(1+acos x)^2}$$
So your integral is $2F(-a)$. First off we will preform the tangent-half-angle substitution $t=tan(x/2)$. Hence we have that
$$F(a)=2int_0^infty frac1{(1+afrac{1-t^2}{1+t^2})^2}frac{mathrm dt}{1+t^2}$$
$$F(a)=2int_0^{infty}frac{1+t^2}{[(1-a)t^2+a+1]^2}mathrm dt$$
Which we can rewrite as
$$F(a)=frac2{1-a}int_0^inftyfrac{mathrm dt}{(1-a)t^2+a+1}+frac{4a}{a-1}int_0^{infty}frac{mathrm dt}{[(1-a)t^2+a+1]^2}$$

Next we consider the very general integral
$$I(m;a,b)=int_0^infty frac{mathrm dx}{(ax^2+b)^{m+1}}$$
First we integrate by parts with $mathrm dv=mathrm dx$ to produce
$$I(m;a,b)=frac{x}{(ax^2+b)^{m+1}}bigg|_0^{infty}+2(m+1)int_0^inftyfrac{ax^2}{(ax^2+b)^{m+2}}mathrm dx$$
$$I(m;a,b)=2(m+1)int_0^inftyfrac{ax^2+b-b}{(ax^2+b)^{m+2}}mathrm dx$$
$$I(m;a,b)=2(m+1)I(m;a,b)-2(m+1)bI(m+1;a,b)$$
Then solving for $I(m+1;a,b)$ and replacing $m+1$ with $m$,
$$I(m;a,b)=frac{2m-1}{2bm}I(m-1;a,b)$$
And for the base case:
$$I(0;a,b)=int_0^{infty}frac{mathrm dx}{ax^2+b}$$
The Trig sub $x=sqrt{frac{b}a}tan u$ gives
$$I(0;a,b)=fracpi{2sqrt{ab}}$$
Which is a special case of
$$int_{x_1}^{x_2}frac{mathrm dx}{ax^2+bx+c}=frac2{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}}bigg|_{x_1}^{x_2}$$
Anyway, the recurrence has the solution
$$I(m;a,b)=fracpi{2^{2m+1}b^msqrt{ab}}{2mchoose m}$$
Which we will apply very shortly!

Recalling the definition of $I(m;a,b)$, we have that
$$F(a)=frac2{1-a}I(0;1-a,1+a)+frac{4a}{a-1}I(1;1-a,1+a)$$
$$F(a)=fracpi{(1-a^2)^{3/2}}$$
And since your integral is given by $2F(-a)$,
$$int_0^{2pi}frac{mathrm dx}{(1-acos x)^2}=frac{2pi}{(1-a^2)^{3/2}}$$

ADDENDUM

Consider the integral
$$C(n;a)=int_0^pi frac{mathrm dx}{(1+acos x)^n}$$
Starting with $t=tanfrac{x}2$,
$$C(n;a)=2int_0^infty frac1{left[1+afrac{1-t^2}{1+t^2}right]^n}frac{mathrm dt}{1+t^2}$$
$$C(n;a)=2int_0^infty frac{left(t^2+1right)^{n-1}}{left[(1-a)t^2+1+aright]^n}mathrm dt$$
Then using the binomial theorem,
$$C(n;a)=2sum_{k=0}^{n-1}{n-1choose k}int_0^inftyfrac{x^{2k}}{left[(1-a)x^2+1+aright]^n}mathrm dx$$
$$C(n;a)=frac2{(1-a)^n}sum_{k=0}^{n-1}{n-1choose k}int_0^inftyfrac{x^{2k}}{left[x^2+frac{1+a}{1-a}right]^n}mathrm dx$$
We then recall the integral due to my collaborator @DavidG,
$$int_0^inftyfrac{x^{q}}{left[x^w+bright]^p}mathrm dx=frac{b^{frac{1+q}{w}-p}}{w}frac{Gammaleft(p-frac{1+q}{w}right)Gammaleft(frac{1+q}{w}right)}{Gammaleft(pright)}$$
Here $Gamma(s)$ is the Gamma function. So with $w=2$, $b=frac{1+a}{1-a}$, $p=n$, and $q=2k$,
$$C(n;a)=frac1{(1-a)^n}sum_{k=0}^{n-1}{n-1choose k}left(frac{1+a}{1-a}right)^{frac{2k+1}{2}-n}frac{Gammaleft(n-frac{2k+1}{2}right)Gammaleft(frac{2k+1}{2}right)}{Gammaleft(nright)}$$
$$C(n;a)=frac1{(1-a)^n}sum_{k=0}^{n-1}left(frac{1+a}{1-a}right)^{frac{2k+1}{2}-n}frac{Gammaleft(n-frac{2k+1}{2}right)Gammaleft(frac{2k+1}{2}right)}{k!Gamma(n-k)}$$