Answer
$(x) \cos^{-1}(\dfrac{x}{2})-\sqrt {4-x^2}+c$
Work Step by Step
The integration by parts formula suggests that $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$
Now, we have
$I=\int \cos^{-1}(\dfrac{x}{2})dx= \cos^{-1}(\dfrac{x}{2})\int dx-\int (\dfrac{x}{2\sqrt {1-(\dfrac{x}{2})^2}} dx$
or, $I=\cos^{-1}(\dfrac{x}{2})\int dx-\int (\dfrac{x}{2\sqrt {1-(\dfrac{x}{2})^2}} dx$
or, $I=(x) \cos^{-1}(\dfrac{x}{2})\int \dfrac{x}{\sqrt {4-x^2}} dx$
consider $p=4-x^2$ or, $-\dfrac{dp}{2}=x dx$
Thus, $I=x \cos^{-1}(\dfrac{x}{2})-\int \dfrac{da}{2 \sqrt a}$
or, $I=(x) \cos^{-1}(\dfrac{x}{2})-\sqrt {4-x^2}+c$
