integral of (x^3-2x+1)sin(4x)

解

\int (x^{3} - 2 x + 1) sin (4 x) d x

- \frac{1}{4} x^{3} cos (4 x) + \frac{3}{16} x^{2} sin (4 x) + \frac{19}{32} x cos (4 x) - \frac{19}{128} sin (4 x) - \frac{1}{4} cos (4 x) + C

解答ステップ

\int (x^{3} - 2 x + 1) sin (4 x) d x

拡張

(x^{3} - 2 x + 1) sin (4 x) : x^{3} sin (4 x) - 2 x sin (4 x) + sin (4 x)

= \int x^{3} sin (4 x) - 2 x sin (4 x) + sin (4 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 (4 x) d x - \int 2 x sin (4 x) d x + \int sin (4 x) d x

\int x^{3} sin (4 x) d x = - \frac{1}{4} x^{3} cos (4 x) + \frac{3}{128} (8 x^{2} sin (4 x) + 4 x cos (4 x) - sin (4 x))

\int 2 x sin (4 x) d x = 2 (- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x))

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

= - \frac{1}{4} x^{3} cos (4 x) + \frac{3}{128} (8 x^{2} sin (4 x) + 4 x cos (4 x) - sin (4 x)) - 2 (- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x)) - \frac{1}{4} cos (4 x)

簡素化

- \frac{1}{4} x^{3} cos (4 x) + \frac{3}{128} (8 x^{2} sin (4 x) + 4 x cos (4 x) - sin (4 x)) - 2 (- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x)) - \frac{1}{4} cos (4 x) : - \frac{1}{4} x^{3} cos (4 x) + \frac{3}{16} x^{2} sin (4 x) + \frac{19}{32} x cos (4 x) - \frac{19}{128} sin (4 x) - \frac{1}{4} cos (4 x)

= - \frac{1}{4} x^{3} cos (4 x) + \frac{3}{16} x^{2} sin (4 x) + \frac{19}{32} x cos (4 x) - \frac{19}{128} sin (4 x) - \frac{1}{4} cos (4 x)

定数を解答に追加する

= - \frac{1}{4} x^{3} cos (4 x) + \frac{3}{16} x^{2} sin (4 x) + \frac{19}{32} x cos (4 x) - \frac{19}{128} sin (4 x) - \frac{1}{4} cos (4 x) + C