Integrate 2 x with respect to x.

The identity is $\sin^2 x = \frac{1 - \cos(2x)}{2}$.
Substitute this into the integral:

$$\int \sin^2 x \,dx = \int \frac{1 - \cos(2x)}{2} \,dx$$

The integral of 1 with respect to $x$ is $x$.
For $\int \cos(2x) \,dx$, use a substitution $u = 2x$, so $du = 2 \,dx$, which means $dx = \frac{1}{2} du$.

$$\int \cos(2x) \,dx = \int \cos(u) \frac{1}{2} \,du = \frac{1}{2} \int \cos(u) \,du = \frac{1}{2} \sin(u) + C_1 = \frac{1}{2} \sin(2x) + C_1$$

$$\frac{1}{2} \left( x - \frac{1}{2} \sin(2x) \right) + C$$ $$= \frac{1}{2}x - \frac{1}{4}\sin(2x) + C$$

The final answer is $\frac{1}{2}x - \frac{1}{4}\sin(2x) + C$.