Answer
$\dfrac{x}{2}-\sin (\dfrac{x}{2} )+C$
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 $(1-\cos^2 x)=\sin^2 x$
Then, we get $\int \sin^2 (\dfrac{x}{4}) dx= \int \dfrac{1-\cos (\dfrac{x}{2})}{2}$
$\implies (\dfrac{1}{2}) x- \int \dfrac{\cos (x/2)}{2}dx=(\dfrac{1}{2}) x-\dfrac{1}{2} [(1/(\dfrac{1}{2} ) \sin (\dfrac{x}{2} )]+C$
Thus, $\int \sin^2 (\dfrac{x}{4} ) dx=\dfrac{x}{2}-\sin (\dfrac{x}{2} )+C$
