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We can solve this using substitution.Let u = sin x.Then du = cos x dx.Substitute these into the integral: ∫ u du = u²/2 + C.Substitute back u = sin x:(sin² x) / 2 + C.Alternatively, we could use the substitution u = cos x. Then du = -sin x dx, so ∫ -u du = -u²/2 + C = -(cos² x)/2 + C. Note that sin² x / 2 and -cos² x / 2 differ by a constant (since sin² x + cos² x = 1, sin² x / 2 = (1 - cos² x) / 2 = 1/2 - cos² x / 2), so both are valid antiderivatives.