$X$ and $Y$ are independent rv having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$ determine pdf of $Z=\frac{X+Y}{3}$ 
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.

So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$. 
We shall find the CDF of $Z$ and then differentiate it.
$F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \  dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
How do I do this? Or otherwise is there different more elegant way to get the required pdf?

$X$ and $Y$ are standard Cauchy hence $\frac{X+Y}{2}$ is standard Cauchy (what easily can be seen using characteristic functions)
So the pdf of $\frac{3}{2}Z$ is also $$f(t) = \frac{1}{\pi}\frac{1}{1+t^2}$$
Hence the distribution of $Z$ is: $$f(z) = \frac{d}{dz}P(Z \le z) = \frac{d}{dz}F\left(\frac{3}{2}Z \le \frac{3}{2}z\right) = \frac{3}{2}f\left(\frac{3}{2}z\right)$$ 
