Answer
$(\dfrac{1}{2}) x-(\dfrac{1}{2}) \sin x+C$
Work Step by Step
Use formula $\int x^{a} dx=\dfrac{x^{a+1}}{n+1}+c$
where $c$ is a constant of proportionality.
As we know that $\cos^2 (x)=\dfrac{1+\cos x}{2}$ and $\int \cos (x) dx=\sin (x)$
Then $\int \cos^2 (\dfrac{x}{2}) dx= \int \dfrac{1+\cos (x)}{2} dx$
This implies that, $(\dfrac{1}{2}) x- \dfrac{1}{2} \int \cos (x) dx=\int \cos^2 (\dfrac{x}{2}) dx=(\dfrac{1}{2}) x-(\dfrac{1}{2}) \sin x+C$
