integral of (x^2-2x+5)cos(3x)

解

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

\frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) - \frac{2}{3} x sin (3 x) - \frac{2}{9} cos (3 x) + \frac{43}{27} sin (3 x) + C

解答ステップ

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

拡張

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

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

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

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

\int 5 cos (3 x) d x = \frac{5}{3} sin (3 x)

= \frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) - \frac{2}{9} (3 x sin (3 x) + cos (3 x)) + \frac{5}{3} sin (3 x)

簡素化

\frac{1}{27} (9 x^{2} sin (3 x) - 2 (- 3 x cos (3 x) + sin (3 x))) - \frac{2}{9} (3 x sin (3 x) + cos (3 x)) + \frac{5}{3} sin (3 x) : \frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) - \frac{2}{3} x sin (3 x) - \frac{2}{9} cos (3 x) + \frac{43}{27} sin (3 x)

= \frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) - \frac{2}{3} x sin (3 x) - \frac{2}{9} cos (3 x) + \frac{43}{27} sin (3 x)

定数を解答に追加する

= \frac{1}{3} x^{2} sin (3 x) + \frac{2}{9} x cos (3 x) - \frac{2}{3} x sin (3 x) - \frac{2}{9} cos (3 x) + \frac{43}{27} sin (3 x) + C