Answer
$\dfrac{x}{2}-\sin (x/2)+C$
Work Step by Step
Calculate the anti-derivative .
Since, we know $1-\cos^2 x=\sin^2 x$
Then $\int \sin^2 (x/4) dx= \int \dfrac{1-\cos (x/2)}{2}$
or, $=\dfrac{1}{2} x- \int \dfrac{\cos (x/2)}{2}dx$
or, $=\dfrac{1}{2} x-\dfrac{1}{2} [(1/(1/2) \sin (x/2)]+C$
Thus, $\int \sin^2 (x/4) dx=\dfrac{x}{2}-\sin (x/2)+C$
