Find the integral
\[ \int \sin^2 x \; dx \]
Use the trigonometric identity \( \; \sin^2 x = \dfrac{1}{2} (1 – \cos (2x)) \) to write
\[ \int \sin^2 x \; dx = \dfrac{1}{2} \int (1 - \cos (2x)) \; dx\]
Apply the sum rule of integrals \( \quad \displaystyle \int (f(x) + g(x) ) dx = \int f(x) dx + \int g(x) dx \) to rewrite the integral as
\[ \int \sin^2 x \; dx = \dfrac{1}{2} \int dx - \int \cos (2x)) \; dx \]
Use the common integrals \( \displaystyle \int \; dx = x \) and \( \displaystyle \int \cos (2x) dx = \dfrac{1}{2} \sin (2x) \) to write the final result as
\[ \boxed { \int \sin^2 x \; dx = \dfrac{1}{2} x - \dfrac{1}{4} \sin (2x) + c } \]
