Let $X$ be a random variable with Cauchy distribution, compute the density function of $Y=\frac{1}{1+X^2}$ 
Let $X$ be a random variable with Cauchy distribution, if $$Y=\frac{1}{1+X^2}$$ then compute the density function of $Y$

I have tried to use some random variable transformation theorem using the functions $\frac{1}{1+X}$ and $X^2$ but I have not been able to get very far, any suggestion or help would be greatly appreciated.

$$P(Y \leq y)=P(X^{2} \geq \frac  1y -1)$$ $$=2P(X \geq \sqrt {\frac 1 y -1})$$ $$=2\int_{\sqrt {\frac 1 y -1}}^{\infty} \frac 1 {\pi} \frac 1 {1+t^{2}} dt.$$ Now just differentiate w.r.t. $y$.
