∫ cos 3 x d x = sin x − 1 3 sin 3 x + C {\displaystyle \int \cos ^{3}xdx=\sin {x}-{\frac {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}{\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 ) {\displaystyle 1-\cos ^{2}(x)=\sin ^{2}(x)}
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9bbe5a17a0d8433bacb3e7c639517a7a91393742)
