integral of (x^2+3)sin(4x)

解

\int (x^{2} + 3) sin (4 x) d x

- \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x) + C

解答ステップ

\int (x^{2} + 3) sin (4 x) d x

部分積分を適用する

= - \frac{1}{4} cos (4 x) (x^{2} + 3) - \int - \frac{1}{2} x cos (4 x) d x

\int - \frac{1}{2} x cos (4 x) d x = - \frac{1}{32} (4 x sin (4 x) + cos (4 x))

= - \frac{1}{4} cos (4 x) (x^{2} + 3) - (- \frac{1}{32} (4 x sin (4 x) + cos (4 x)))

簡素化

- \frac{1}{4} cos (4 x) (x^{2} + 3) - (- \frac{1}{32} (4 x sin (4 x) + cos (4 x))) : - \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x)

= - \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x)

定数を解答に追加する

= - \frac{1}{4} x^{2} cos (4 x) + \frac{1}{8} x sin (4 x) - \frac{23}{32} cos (4 x) + C