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