This question was also asked [here](https://math.stackexchange.com/questions/2887659/i-really-dont-know-what-else-i-can-do-to-solve-this-integral) and [here](https://mathematica.stackexchange.com/questions/180192/why-cant-mathematica-compute-this-integral-int-0-pi-2-frac1-a-cos2x).


I have faced some difficulties to do the following integral:



> 
> $$ I=\int\_{0}^{2\pi}d\phi\int\_{0}^{\pi}d\theta~\sin\theta\int\_{0}^{\infty}dr~r^2\frac{3x^2y^2\cos(u r \sin\theta \cos\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{r^2}{2}} \tag{1}, $$
> 
> 
> 


where $x$, $y$, and $u$ are real positive constants. I tried at least two ways to solve this integral:


* First attempt:


I began to solve the $r$ integral first. By using Mathematica, then


$$ I=\int\_{0}^{2\pi}d\phi\int\_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2(1-u^2\sin^2\theta\cos^2\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{u^2}{2}\sin^2\theta\cos^2\phi} \tag{2}. $$


After that, I looked for a solution for $\phi$ integral. My best attempt was:


$$ I\_\phi(x,y,u,\theta)=\frac{2}{B}\left[B\left(\frac{1}{2}\right)F\_1\left(\frac{1}{2},1,-;1;\nu,-\frac{a}{2}\right)-aB\left(\frac{1}{2}\right)F\_1\left(\frac{3}{2},1,-;2;\nu,-\frac{a}{2}\right)\right], $$


where $B=x^2\sin^2\theta+x^2y^2\cos^2\theta$, $a=u^2\sin^2\theta$, and $\nu=\frac{x^2-y^2}{x^2+x^2y^2\cot^2\theta}$. In this way, the final results it's something like that:


$$ I= \int\_{0}^{\pi} \mathrm d \theta~3x^2y^2\sin\theta \cos^2\theta~ I\_\phi(x,y,u,\theta). \tag{3}. $$ 


Eq. $(3)$ cannot be further simplied in general and is the nal result.




---


* Second attempt:


To avoid the hypergeometric function $F\_1$, I tried to start with the $\phi$ integral. In this case, my initial problem is an integral something like that:


$$ \int\_{0}^{2\pi} \mathrm d \phi \frac{\cos(A \cos\phi)}{a^2\cos^2\phi+b^2\sin^2\phi}. \tag{4} $$


This integral $(4)$ can be solved by series (see [Vincent's answer](https://math.stackexchange.com/questions/2886810/evaluation-of-int-0-pi-2-frac-cos-left-a-cos-t-righta2) and [Jack's answer](https://math.stackexchange.com/questions/2887659/i-really-dont-know-what-else-i-can-do-to-solve-this-integral/2887771#2887771)). However those solutions, at least for me, has not a closed form. This is my final step on this second attempt :(




---


What is the point? It turns out that someone has managed to solve the integral $(1)$, at least the integral in $r$ and $\phi$. The final resuls found by this person was:



> 
> $$ I\_G=\frac{12 \pi x~y}{(1-x^2)^{3/2}}\int\_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2 \exp\left(-\frac{u^2}{2}\frac{x^2k^2}{(1-x^2)(1-k^2)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}, $$
> 
> 
> 


where, I belive, $k=\sqrt{1-x^2}\cos\theta$. As you can see in this following code performed in Mathematica



```
IG[x_, y_, u_] := 
     Sqrt[Pi/2] NIntegrate[(12  Pi x y)/(1 - x^2)^(3/2)
    (v^2 Exp[-(u^2 x^2 v^2)/(2 (1 - x^2) (1 - v^2))])/(Sqrt[1 - v^2] Sqrt[1 - v^2 (1 - y^2)/(1 - x^2)]), {v, 0, Sqrt[1 - x^2]}]
    IG[.3, .4, 1]
    ** 4.53251 **

I[x_, y_, u_] := 
 NIntegrate[(r^2 Sin[a] Cos[
      u r Sin[a] Cos[b]] 3 x^2 y^2 Cos[a]^2 Exp[-r^2/
       2])/((y^2 Cos[b]^2 + x^2 Sin[b]^2) Sin[a]^2 + 
     x^2 y^2 Cos[a]^2), {r, 0, Infinity}, {a, 0, Pi}, {b, 0, 2 Pi}]
I[.3, .4, 1]
** 4.53251 **

```

the integrals $I$ and $I\_G$ are equals. Indeed, since that they emerge from the same physical problem. 


**So, my question is: what are the steps applied for that integral $I$ gives the integral $I\_G$?**




---


**Edit**


Since my question was not solved yet, I think it is because it is a tough question, I will show a particular case of the integral $I$, letting $u=0$. I hope with this help you help me. 


In this case, the $r$ integral in $(1)$ is trivial and the integral takes the form:


$$ I\_P=\int\_{0}^{2\pi}d\phi\int\_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}. \tag{5} $$


The $\phi$ integral can be integrated with the help of Eq. 3.642.1 in Gradstein and Ryzhik's tables of integrals. Thereby, the $I\_P$ takes the for:


$$ I\_P=3xy\int\_{0}^{\pi}d\theta\frac{\sin\theta\cos^2\theta}{\sqrt{1+(x^2-1)\cos^2\theta}\sqrt{1+(y^2-1)\cos^2\theta}}. \tag{6}$$


Now the change of variable $k=\sqrt{1-x^2}\cos\theta$ bring expression $(6)$ to the form


$$ I\_P= \frac{(const) x~y}{(1-x^2)^{3/2}}\int\_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}. $$


Did you notice how $I\_G$ and $I\_P$ are similar? Do you think a similar approach can be applied to my original problem? Please, let me know.




---




---



> 
> I've solved this problem applying the Schwinger proper-time substitution: $$\frac{1}{q^2}=\int\_{0}^{\infty}\mathrm{d\xi}~\mathrm{e^{-q^2\xi}}. $$
> 
> 
> 



Difficult trigonometric integral
Here is an outline of the approach I have taken to solve this integral. 


First rewrite the integral $(1)$ in Cartesian variables:


$$I=\int\_{-\infty}^{\infty} \mathrm{d}^3v~ \frac{3x^2y^2v\_z^2}{y^2v\_x^2+x^2v\_y^2+x^2y^2v\_z^2}\cos(uv\_x)\exp\left(-\frac{v\_x^2}{2}-\frac{v\_y^2}{2}-\frac{v\_z^2}{2}\right). $$


Now use the following substitution


$$ \frac{1}{y^2v\_x^2+x^2v\_y^2+x^2y^2v\_z^2}=\int\_{0}^{\infty}d\tau~\mathrm{ e^{-(y^2v\_x^2+x^2v\_y^2+x^2y^2v\_z^2)\tau}},$$


such that


$$ I=\int\_{0}^{\infty}d\tau\int\_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v\_z^2\cos(uv\_x) \mathrm{e^{-v\_x^2(\tau y^2+1/2)-v\_y^2(\tau x^2+1/2)-v\_z^2(\tau x^2y^2+1/2)}}. $$


The $(v\_x,v\_y,v\_z)$ can be evaluated with the help of Mathematica. The results gives


$$ \int\_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v\_z^2\cos(uv\_x) \mathrm{e^{-\alpha v\_x^2-\beta v\_y^2-\gamma v\_z^2}}=\frac{3\pi^{3/2}}{2x^2y^2}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}.$$


Thereby, 


$$ I= -\frac{3~\mathrm{const}}{x^2y^2~\mathrm{const}}\int\_{0}^{\infty}\mathrm{d\tau}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}. $$


Now, performing the substitution $\tau=\frac{1-x^2}{2x^2y^2k^2}-\frac{1}{2x^2y^2}$ gives us



> 
> $$ I=(\mathrm{const})~\frac{3x~y}{(1-x^2)^{3/2}}\int\_{0}^{\sqrt{1-x^2}}\mathrm{dk}\frac{k^2\exp\left(-\frac{u^2}{2}\frac{x^2k^2}{\left(1-x^2\right)\left(1-k^2\right)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}, $$
> 
> 
> 


which is the desired integral unless of a constant. ;)


