Answer
$$x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{2}\sqrt {1 - 4{x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx \cr
& {\text{substitute }}u = {\cos ^{ - 1}}\left( {2x} \right),{\text{ }}du = - \frac{2}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}dx \cr
& du = - \frac{2}{{\sqrt {1 - 4{x^2}} }}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{using integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \int {x\left( { - \frac{2}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}} \right)dx} \cr
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) + 2\int {\frac{x}{{\sqrt {1 - 4{x^2}} }}dx} \cr
& {\text{find antiderivative}}{\text{, }}t = 1 - 2{x^2},{\text{ }}dt = - 8xdx \cr
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) + 2\int {\frac{{\left( { - 1/8} \right)dt}}{{\sqrt t }}} \cr
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{4}\int {{t^{ - 1/2}}} dt \cr
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{4}\left( {2{t^{1/2}}} \right) + C \cr
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{2}\sqrt t + C \cr
& {\text{replace }}t = 1 - 4{x^2} \cr
& \int {{{\cos }^{ - 1}}\left( {2x} \right)} dx = x{\cos ^{ - 1}}\left( {2x} \right) - \frac{1}{2}\sqrt {1 - 4{x^2}} + C \cr} $$
