Answer
$ - 2\cos (\dfrac{1}{2} x) +C $
Work Step by Step
Our aim is to solve the integral $ \int \sin (\dfrac{1}{2} x) \ dx$
Use formula:$ \int \sin (ax+b) \ dx =\dfrac{-1}{a} \cos (ax+b)+C$
Let us consider that $a =\dfrac{1}{2}$ and $b=0$
Now, $ \int \sin (\dfrac{1}{2} x) \ dx =\dfrac{-1}{\frac{1}{2}} \cos (\dfrac{1}{2} x) +C $
or, $= - 2\cos (\dfrac{1}{2} x) +C $
