∫ cos 3 x d x = sin x − 1 3 sin 3 x + C d d x [ sin x − 1 3 sin 3 x + C ] = cos x − 1 3 ⋅ 3 sin 2 x cos x + 0 = cos x − sin 2 x cos x = cos x − ( 1 − c o s 2 ( x ) ) cos x = cos 3 x {\displaystyle {\begin{aligned}&\int \cos ^{3}xdx=\sin {x}-{\frac {1}{3}}\sin ^{3}x+C\\[2ex]&{\frac {d}{dx}}{[\sin {x}-{\frac {1}{3}}\sin ^{3}{x}+C]}\\[2ex]&={\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}}}