integral of (x^2+7x-5)cos(2x)

解

\int (x^{2} + 7 x - 5) cos (2 x) d x

\frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) + \frac{7}{2} x sin (2 x) - \frac{11}{4} sin (2 x) + \frac{7}{4} cos (2 x) + C

解答ステップ

\int (x^{2} + 7 x - 5) cos (2 x) d x

拡張

(x^{2} + 7 x - 5) cos (2 x) : x^{2} cos (2 x) + 7 x cos (2 x) - 5 cos (2 x)

= \int x^{2} cos (2 x) + 7 x cos (2 x) - 5 cos (2 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 (2 x) d x + \int 7 x cos (2 x) d x - \int 5 cos (2 x) d x

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

\int 7 x cos (2 x) d x = \frac{7}{4} (2 x sin (2 x) + cos (2 x))

\int 5 cos (2 x) d x = \frac{5}{2} sin (2 x)

= \frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) - \frac{1}{4} sin (2 x) + \frac{7}{4} (2 x sin (2 x) + cos (2 x)) - \frac{5}{2} sin (2 x)

簡素化

\frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) - \frac{1}{4} sin (2 x) + \frac{7}{4} (2 x sin (2 x) + cos (2 x)) - \frac{5}{2} sin (2 x) : \frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) + \frac{7}{2} x sin (2 x) - \frac{11}{4} sin (2 x) + \frac{7}{4} cos (2 x)

= \frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) + \frac{7}{2} x sin (2 x) - \frac{11}{4} sin (2 x) + \frac{7}{4} cos (2 x)

定数を解答に追加する

= \frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) + \frac{7}{2} x sin (2 x) - \frac{11}{4} sin (2 x) + \frac{7}{4} cos (2 x) + C