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\begin{aligned} \int \sin ^{2}(x) \mathrm{dx} &=\int \frac{1-\cos (2 x)}{2} \mathrm{dx}=\int \frac{1}{2} \mathrm{~d} x-\int \frac{1}{2} \cos (2 x) \mathrm{d} x \\ &=\frac{x}{2}-\frac{1}{2} \frac{\sin (2 x)}{2}+\mathrm{C}=\frac{x}{2}-\frac{\sin (2 x)}{4}+\mathrm{C} \text { or } \frac{1}{2}\{x-\sin (x) \cos (x)\}+\mathrm{C} \end{aligned}