integral of (x^3+2x+14)sin(3x)

解

\int (x^{3} + 2 x + 14) sin (3 x) d x

- \frac{1}{3} x^{3} cos (3 x) + \frac{1}{3} x^{2} sin (3 x) - \frac{4}{9} x cos (3 x) + \frac{4}{27} sin (3 x) - \frac{14}{3} cos (3 x) + C

解答ステップ

\int (x^{3} + 2 x + 14) sin (3 x) d x

拡張

(x^{3} + 2 x + 14) sin (3 x) : x^{3} sin (3 x) + 2 x sin (3 x) + 14 sin (3 x)

= \int x^{3} sin (3 x) + 2 x sin (3 x) + 14 sin (3 x) d x

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 x^{3} sin (3 x) d x + \int 2 x sin (3 x) d x + \int 14 sin (3 x) d x

\int x^{3} sin (3 x) d x = - \frac{1}{3} x^{3} cos (3 x) + \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x)))

\int 2 x sin (3 x) d x = 2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x))

\int 14 sin (3 x) d x = - \frac{14}{3} cos (3 x)

= - \frac{1}{3} x^{3} cos (3 x) + \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + 2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x)) - \frac{14}{3} cos (3 x)

簡素化

- \frac{1}{3} x^{3} cos (3 x) + \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) + 2 (- \frac{1}{3} x cos (3 x) + \frac{1}{9} sin (3 x)) - \frac{14}{3} cos (3 x) : - \frac{1}{3} x^{3} cos (3 x) + \frac{1}{3} x^{2} sin (3 x) - \frac{4}{9} x cos (3 x) + \frac{4}{27} sin (3 x) - \frac{14}{3} cos (3 x)

= - \frac{1}{3} x^{3} cos (3 x) + \frac{1}{3} x^{2} sin (3 x) - \frac{4}{9} x cos (3 x) + \frac{4}{27} sin (3 x) - \frac{14}{3} cos (3 x)

定数を解答に追加する

= - \frac{1}{3} x^{3} cos (3 x) + \frac{1}{3} x^{2} sin (3 x) - \frac{4}{9} x cos (3 x) + \frac{4}{27} sin (3 x) - \frac{14}{3} cos (3 x) + C