integral of (3-2x)*sin(5-4x)

解

\int (3 - 2 x) \cdot sin (5 - 4 x) d x

\frac{3}{4} cos (5 - 4 x) - 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x)) + C

解答ステップ

\int (3 - 2 x) sin (5 - 4 x) d x

拡張

(3 - 2 x) sin (5 - 4 x) : 3 sin (5 - 4 x) - 2 x sin (5 - 4 x)

= \int 3 sin (5 - 4 x) - 2 x sin (5 - 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 3 sin (5 - 4 x) d x - \int 2 x sin (5 - 4 x) d x

\int 3 sin (5 - 4 x) d x = \frac{3}{4} cos (5 - 4 x)

\int 2 x sin (5 - 4 x) d x = 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x))

= \frac{3}{4} cos (5 - 4 x) - 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x))

定数を解答に追加する

= \frac{3}{4} cos (5 - 4 x) - 2 (\frac{1}{4} x cos (5 - 4 x) + \frac{1}{16} sin (5 - 4 x)) + C