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

解

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

- x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x) + C

解答ステップ

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

x + 3 = 3 (\frac{x}{3} + 1)

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

簡素化

(3 (\frac{x}{3} + 1))^{2} sin (x) : x^{2} sin (x) + 6 x sin (x) + 9 sin (x)

= \int x^{2} sin (x) + 6 x sin (x) + 9 sin (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} sin (x) d x + \int 6 x sin (x) d x + \int 9 sin (x) d x

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

\int 6 x sin (x) d x = 6 (- x cos (x) + sin (x))

\int 9 sin (x) d x = - 9 cos (x)

= - x^{2} cos (x) + 2 (x sin (x) + cos (x)) + 6 (- x cos (x) + sin (x)) - 9 cos (x)

簡素化

- x^{2} cos (x) + 2 (x sin (x) + cos (x)) + 6 (- x cos (x) + sin (x)) - 9 cos (x) : - x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x)

= - x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x)

定数を解答に追加する

= - x^{2} cos (x) + 2 x sin (x) - 6 x cos (x) - 7 cos (x) + 6 sin (x) + C