How can I show that $ \cos^{-1}(x)=-\int\frac{dx}{\sqrt{1-x^2}}+c? $ 
How can I show that
  $$
\cos^{-1}(x)=-\int\frac{dx}{\sqrt{1-x^2}}+c?
$$

 A: 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: $$\forall x\in]-1,1[,\arccos'(x)=-\frac{1}{\sqrt{1-x^2}}.$$
A: Let $\theta=\cos^{-1}x$. Then $0\le\theta\le\pi$ therefore $0\le\sin\theta\le1$
You want to show that 
\begin{equation}
\cos^{-1}x+\int\frac{dx}{\sqrt{1-x^2}}=c
\end{equation}
for some constant $c$.
Note that $x=\cos{\theta},\,dx=-\sin\theta\,d\theta, \sqrt{1-x^2}=\sin\theta$
Therefore
\begin{eqnarray}
\cos^{-1}x+\int\frac{dx}{\sqrt{1-x^2}}&=&\theta-\int \frac{\sin\theta}{\sin\theta}\\
&=&\theta-\int d\theta\\
&=&\theta-\theta+c\\
&=&c
\end{eqnarray}
A: Let $$x=\cos  u\Rightarrow dx=-\sin u du,u=\cos^{-1}x$$
Then
$$-\int\frac{dx}{\sqrt{1-x^2}}=\int\frac{\sin u \ du}{\sqrt{1-\cos^2u}}=\int\frac{\sin u}{|\sin u|}du=\begin{cases}u+c,& 0\le u \le \pi\\ -u+c,& -\pi<u<0\end{cases}$$
which gives the desired result for $0\le u\le\pi$.
A: Just differentiate the function $x\mapsto \cos(\cos^{-1}(x)) = x$ 
Use the chain rule $(f\circ g)' = g'.f'\circ g.$ 
