upload solution for problem 5.9

Author: RamonAraujo

src/chapters/5/sections/pdfs_and_cdfs/index.tex

@@ -12,3 +12,5 @@ \subsection{problem 6}
 \input{problems/6}
 \subsection{problem 7}
 \input{problems/7}
+\subsection{problem 9}
+\input{problems/9}

src/chapters/5/sections/pdfs_and_cdfs/problems/9.tex

@@ -0,0 +1,17 @@
+Let $F$ be the CDF of the Cauchy distribution.
+
+$$
+F(x) = P(X \le x) = \int_{-\infty}^x f(t) dt = \int_{-\infty}^x \frac{1}{\pi(1+t^2)} dt
+$$
+
+Making the substitution $u = \arctan(t)$, we obtain $du = \frac{dt}{1+t^2}$.
+The lower and upper limits of integration become $-\pi/2$ and $\arctan(x)$, respectively.
+
+$$
+F(x) = \int_{-\pi/2}^{\arctan(x)} \frac{1}{\pi} \, du = \frac{1}{\pi} \left( \arctan(x) + \frac{\pi}{2} \right)
+$$
+
+$$
+F(x) = \frac{1}{2} + \frac{1}{\pi} \arctan(x)
+$$
+