What is the antiderivative of sin^2(x)?

Jul 4, 2016

$= \frac{1}{2} \left[x - \frac{1}{2} \sin 2 x\right] + C$

Explanation:

We're going to use the trig identity

$\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$

$\implies {\sin}^{2} x = \frac{1}{2} \left(1 - \cos 2 x\right)$

So $\int {\sin}^{2} x \mathrm{dx} = \frac{1}{2} \int \left(1 - \cos 2 x\right) \mathrm{dx}$

$= \frac{1}{2} \left[x - \frac{1}{2} \sin 2 x\right] + C$