Answer
$(x) \cos^{-1}(x/2)-\sqrt {4-x^2}+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
$\int (1) \cos^{-1}(x/2)dx= \cos^{-1}(x/2)\int dx-\int (\dfrac{x}{2\sqrt {1-(x/2)^2}} dx$
$\implies \cos^{-1}(x/2)\int dx-\int (\dfrac{x}{2\sqrt {1-(x/2)^2}} dx=(x) \cos^{-1}(x/2)\int \dfrac{x}{\sqrt {4-x^2}} dx$
Plug $a=4-x^2 \implies \dfrac{-da}{2}=x dx$
Then, we have $(x) \cos^{-1}(x/2)-\int \dfrac{da}{2 \sqrt a}=(x) \cos^{-1}(x/2)-\sqrt {4-x^2}+c$
