Banach Contraction mapping of $\Phi(f)(x)=\int_0^x \frac{1}{1+f(t)^2}dt$ , find a fixed point. 
Let $(X,d)$ be metric space with $d(f,g)=\sup |f(x)-g(x)|$ where $X$ is the set of continuous function on $[0,1/2]$. Show $\Phi:X\rightarrow X$ $$\Phi(f)(x)=\int_0^x \frac{1}{1+f(t)^2}dt$$ has a unique fixed point $f(0)=0$.
b) Show that it satisfies $\frac{df}{dx}=\frac{1}{1+f(x)^2}$

My attempt:
I assumed $[0,1/2]$ is complete. I need to show that:
$$d(\Phi(f)(x),\Phi(g)(x))=\sup \Big|\int_0^x \frac{1}{1+f(t)^2}dt-\int_0^x \frac{1}{1+g(t)^2}dt\Big|\leq\alpha d(f,g)$$
Now,
$$\sup \Big|\int_0^x \frac{1}{1+f(t)^2}dt-\int_0^x \frac{1}{1+g(t)^2}dt\Big|=\sup \Big|\int_0^x \frac{1}{1+f(t)^2+g(t)^2}dt\Big|$$
Now ideally I need to somehow change this integral to include $d(f,g)=\sup|f(x)-g(x)|$, but I don't know how.
Then I would integrate, and get the integration constant to be $<1$, which will be my contraction constant.

I'm unable to comment due to low reputation, so I'll have to use this instead.
Have you taken a look at Contraction Mapping question?
