Let X be the set of continuous real valued functions defined on $[0,\frac{1}{2}]$ with the metric $d(f,g):=\sup_{x\in[0,\frac{1}{2}]} |f(x)-g(x)|$.
Define the map $\theta:X\rightarrow X$ such that $\theta (f)(x)=\int_{0}^{x} \frac{1}{1+f(t)^2} dt$.
I need to show that $\theta$ is a contraction mapping and that the unique fixed point satisfies $f(0)=0$ and the differential equation $\frac{df}{dx}=\frac{1}{1+f(x)^2}$
So I'm pretty lost on this, I'm quite comfortable proving that things like $f(x)=1+\frac{1}{1+x^4}$ are contraction mapping but I'm a bit confused with this, so
$d(\theta(f),\theta(g))=\sup|\int_{0}^{x} \frac{1}{1+f(t)^2}-\frac{1}{1+g(t)^2} dt|$
and so this is: $=\sup|\int_{0}^{x} \frac{(g(t)-f(t))(g(t)+f(t))}{(1+f(t)^2)(1+g(t)^2)} dt|$
but I am unsure where to go from here, I know that I need to get this to be something like:
$\leq\alpha\sup|f(x)-g(x)|$ where $\alpha$ is the contraction constant?
Thanks very much for any help