Answer
$$\int arccos\,x\,dx=x\,arccos\,x-\sqrt{1-x^{2}}+C$$
Work Step by Step
${(x\,arccos\,x)}'=arccos\,x+\frac{x}{\sqrt{1-x^{2}}}$
$$\int arccos\,x\,dx=x\,arccos\,x+\int \frac{x}{\sqrt{1-x^{2}}}dx$$
$For\ \int \frac{x}{\sqrt{1-x^{2}}}dx,\ let\ x=sin\,t,dx=cos\,t\,dt.$
$$\int \frac{x}{\sqrt{1-x^{2}}}dx=\int \frac{sin\,t}{cos\,t}cos\,t\,dt$$
$$=-cos\,t+C=-\sqrt{1-x^{2}}+C$$
$$\int arccos\,x\,dx=x\,arccos\,x-\sqrt{1-x^{2}}+C$$
