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

解

\int (2 - x^{2}) sin (2 x + 1) d x

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

解答ステップ

\int (2 - x^{2}) sin (2 x + 1) d x

部分積分を適用する

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

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

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

簡素化

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

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

定数を解答に追加する

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