How can I show that $ \cos^{-1}(x)=-\int\frac{dx}{\sqrt{1-x^2}}+c? $ How can I show that
  $$
\cos^{-1}(x)=-\int\frac{dx}{\sqrt{1-x^2}}+c?
$$
One has: $$\forall x\in]-1,1[,(\cos\circ\arccos)(x)=x.$$
Therefore, one gets: $$\forall x\in ]-1,1[,-\arccos'(x)\sin(\arccos(x))=1.$$
For all $y\in\mathbb{R}$, $\sin^2(y)=1-\cos^2(y)$, using this identity for $y=\arccos(x)$ with $x\in ]-1,1[$ and noticing that $\sin\geqslant 0$ on $]0,\pi[\ni\arccos(x)$, one has: $$\sin(\arccos(x))=\sqrt{1-x^2}.$$
Finally, one has: $$\forall x\in]-1,1[,\arccos'(x)=-\frac{1}{\sqrt{1-x^2}}.$$
Just differentiate the function $x\mapsto \cos(\cos^{-1}(x)) = x$ 

Use the chain rule $(f\circ g)' = g'.f'\circ g.$