5.4 Indefinite Integrals and the Net Change Theorem/3

$\int \cos ^{3}xdx=\sin {x}-{\frac {1}{3}}\sin ^{3}x+C$

${\begin{aligned}{\frac {d}{dx}}{[\sin {x}-{\frac {1}{3}}\sin ^{3}{x}+C]}&={\cos {x}-{\frac {1}{3}}\cdot 3\sin ^{2}{x}\cos {x}+0}\\[2ex]&=\cos {x}-\sin ^{2}{x}\cos {x}\\[2ex]&=\cos {x}-(1-cos^{2}{x})\cos {x}\\[2ex]&=\cos ^{3}{x}\end{aligned}}$

Note: $1-\cos ^{2}(x)=\sin ^{2}(x)$