Integral of $$\sin^2 x$$

 

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 $\boxed { \int \sin^2 x \; dx = \dfrac{1}{2} x - \dfrac{1}{4} \sin (2x) + c }$
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