# Integral of $\sin^4(x)$

 

Evaluate the integral $\int \sin^4(x) \; dx$ Rewrite as $= \int \sin^2(x) \sin^2(x) \; dx$
Use the
trigonometric identity $\sin^2 x = \dfrac{1}{2} (1 - \cos (2 x) )$ to rewrite as
$= \dfrac{1}{4} \int (1 - \cos (2 x) ) (1 - \cos (2 x) ) \; dx$
Expand the the integrand $= \dfrac{1}{4} \int ( 1 - 2 \cos (2) + \cos^2 (2 x) ) \; dx$ Use
trigonometric identity $\cos^2 x = \dfrac{1}{2} (1 + \cos (2 x) )$ to change $\cos^2 (2 x)$ included in the integrand $= \dfrac{1}{4} \int ( 1 - 2 \cos (2 x) + \dfrac{1}{2} (1 + \cos (4 x) ) ) \; dx$ Simplify $= \dfrac{1}{4} \int ( \dfrac{3}{2} - 2 \cos (2 x) + \dfrac{1}{2} \cos (4 x) ) \; dx$ Use the common integral $\int \cos (kx) \; dx = \dfrac{1}{k} \sin (kx) + c$ to evaluate the given integral $\boxed {\int \sin^4(x) \; dx = \dfrac{3}{8} x - \dfrac{1}{4} \sin (2x) + \dfrac{1}{32} \sin (4x) + c}$