question about integrals and calculus ?

No; we are dealing strictly with a squared trig function, so it doesn't work like a squared variable. Here is how we solve it.

Integral ( cos^2(x/2) dx )

We cannot solve this directly. We must use the half angle identity:

cos^2(y) = (1/2)(1 + cos(2y))

In this case, y = (x/2), so

cos^2(x/2) = (1/2)(1 + cos(2*(x/2)))

cos^2(x/2) = (1/2)(1 + cos(x))

Using that identity, our integral becomes

Integral ( (1/2)(1 + cos(x)) dx )

(1/2) Integral ( (1 + cos(x)) dx )

And it becomes trivial.

(1/2)(x + sin(x)) + C

(1/2)x + (1/2)sin(x) + C

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If we WERE to take the derivative of the suggested answer, we would not get the original question. Let me show you.

f(x) = (1/3) [ sin(x/2) ]^3

We use the power rule (so the 3 cancels out the (1/3), and then the chain rule (twice).

f'(x) = [sin(x/2)]^2 cos(x/2) (1/2)

Looks nothing like the original question.