\(\int sin^4xdx=\int\left(\dfrac{1-cos2x}{2}\right)^2dx=\dfrac{1}{4}\int\left(cos^22x-2cos2x+1\right)dx\)
\(=\dfrac{1}{4}\int\left(\dfrac{1}{2}+\dfrac{1}{2}cos4x-2cos2x+1\right)dx=\int\left(\dfrac{3}{8}+\dfrac{cos4x}{8}-\dfrac{cos2x}{2}\right)dx\)
\(=\dfrac{3x}{8}+\dfrac{sin4x}{32}-\dfrac{sin2x}{4}+C\)