Consider the following:
Let $X$, $Y$ be two jointly continuous random variables with joint PDF $$f\left(x,y\right)=\frac{c}{2\pi}\frac{1}{\left(c^{2}+x^{2}+y^{2}\right)^{\frac{3}{2}}}\quad\text{(Standard Bivariate Cauchy Distribution)}.$$ Find the marginal PDF's of X and Y.
I know, I checked with Mathematica, the marginal PDF's of $X$ and $Y$ to be $$f_{X}(x)=2\int_{0}^{\infty}\frac{c}{2\pi}\frac{1}{\left(c^{2}+x^{2}+y^{2}\right)^{\frac{3}{2}}}dy=\frac{c}{\pi(c^{2}+x^{2})}$$ ad $$f_{Y}(y)=2\int_{0}^{\infty}\frac{c}{2\pi}\frac{1}{\left(c^{2}+x^{2}+y^{2}\right)^{\frac{3}{2}}}dx=\frac{c}{\pi(c^{2}+y^{2})}$$ but how? I mean I can't even begin to imagine how one computes those integrals
Okay, I won't calculate exactly those integrals, but something which would be sufficient.
You need to calculate integral of type ($a>0$)$$\int_0^\infty \frac{dx}{(a+x^2)^{\frac{3}{2}}}$$
Substitute $t=\frac{x}{\sqrt{a}}$ you get $$\frac{\sqrt{a}}{a^{\frac{3}{2}}} \int_0^\infty \frac{1}{(1+t^2)^{\frac{3}{2}}}dt$$
So we'll just calculate $$ \int_0^\infty \frac{1}{(1+t^2)^{\frac{3}{2}}} dt$$
And when one see $t^2+1$, one should think about $t=\tan(u)$ substitution, getting: $$ \int_0^\frac{\pi}{2} \frac{1+\tan(u)^2}{(\tan(u)^2+1)^{\frac{3}{2}}} du = \int_0^\frac{\pi}{2} \cos(u)du = 1$$
By that our first integral is equal to $\frac{1}{a}$
In your case it gives $$\frac{c}{\pi}\int_0^\infty \frac{1}{(c^2+z^2 + t^2)^{\frac{3}{2}}}dt = \frac{c}{\pi(c^2+z^2)}$$ where either $(z,t) = (x,y)$ or $(z,t)=(y,x)$