integral of-cos(u)+1/3 cos^3(u)

解

\int - cos (u) + \frac{1}{3} cos^{3} (u) d u

- \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u) + C

解答ステップ

\int - cos (u) + \frac{1}{3} cos^{3} (u) d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int cos (u) d u + \int \frac{1}{3} cos^{3} (u) d u

\int cos (u) d u = sin (u)

\int \frac{1}{3} cos^{3} (u) d u = \frac{1}{3} (sin (u) - \frac{sin ^{3} ( u )}{3})

= - sin (u) + \frac{1}{3} (sin (u) - \frac{sin ^{3} ( u )}{3})

簡素化

- sin (u) + \frac{1}{3} (sin (u) - \frac{sin ^{3} ( u )}{3}) : - \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)

= - \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u)

定数を解答に追加する

= - \frac{2}{3} sin (u) - \frac{1}{9} sin^{3} (u) + C