integral of (2-x)sin(x+1)

解

\int (2 - x) sin (x + 1) d x

- 2 cos (x + 1) + x cos (x + 1) - sin (x + 1) + C

解答ステップ

\int (2 - x) sin (x + 1) d x

拡張

(2 - x) sin (x + 1) : 2 sin (x + 1) - x sin (x + 1)

= \int 2 sin (x + 1) - x sin (x + 1) 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 2 sin (x + 1) d x - \int x sin (x + 1) d x

\int 2 sin (x + 1) d x = - 2 cos (x + 1)

\int x sin (x + 1) d x = - x cos (x + 1) + sin (x + 1)

= - 2 cos (x + 1) - (- x cos (x + 1) + sin (x + 1))

簡素化

= - 2 cos (x + 1) + x cos (x + 1) - sin (x + 1)

定数を解答に追加する

= - 2 cos (x + 1) + x cos (x + 1) - sin (x + 1) + C