integral of (x^3-5x+10)*sin(2x)

解

\int (x^{3} - 5 x + 10) \cdot sin (2 x) d x

- \frac{1}{2} x^{3} cos (2 x) + \frac{3}{4} x^{2} sin (2 x) + \frac{13}{4} x cos (2 x) - \frac{13}{8} sin (2 x) - 5 cos (2 x) + C

解答ステップ

\int (x^{3} - 5 x + 10) sin (2 x) d x

拡張

(x^{3} - 5 x + 10) sin (2 x) : x^{3} sin (2 x) - 5 x sin (2 x) + 10 sin (2 x)

= \int x^{3} sin (2 x) - 5 x sin (2 x) + 10 sin (2 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 (2 x) d x - \int 5 x sin (2 x) d x + \int 10 sin (2 x) d x

\int x^{3} sin (2 x) d x = - \frac{1}{2} x^{3} cos (2 x) + \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

\int 5 x sin (2 x) d x = 5 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))

\int 10 sin (2 x) d x = - 5 cos (2 x)

= - \frac{1}{2} x^{3} cos (2 x) + \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) - 5 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) - 5 cos (2 x)

簡素化

- \frac{1}{2} x^{3} cos (2 x) + \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)) - 5 (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) - 5 cos (2 x) : - \frac{1}{2} x^{3} cos (2 x) + \frac{3}{4} x^{2} sin (2 x) + \frac{13}{4} x cos (2 x) - \frac{13}{8} sin (2 x) - 5 cos (2 x)

= - \frac{1}{2} x^{3} cos (2 x) + \frac{3}{4} x^{2} sin (2 x) + \frac{13}{4} x cos (2 x) - \frac{13}{8} sin (2 x) - 5 cos (2 x)

定数を解答に追加する

= - \frac{1}{2} x^{3} cos (2 x) + \frac{3}{4} x^{2} sin (2 x) + \frac{13}{4} x cos (2 x) - \frac{13}{8} sin (2 x) - 5 cos (2 x) + C