integral of (x^2+1)*sin(3x)

解

\int (x^{2} + 1) \cdot sin (3 x) d x

- \frac{1}{3} x^{2} cos (3 x) + \frac{2}{9} x sin (3 x) - \frac{7}{27} cos (3 x) + C

解答ステップ

\int (x^{2} + 1) sin (3 x) d x

部分積分を適用する

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

\int - \frac{2}{3} x cos (3 x) d x = - \frac{2}{27} (3 x sin (3 x) + cos (3 x))

= - \frac{1}{3} cos (3 x) (x^{2} + 1) - (- \frac{2}{27} (3 x sin (3 x) + cos (3 x)))

簡素化

- \frac{1}{3} cos (3 x) (x^{2} + 1) - (- \frac{2}{27} (3 x sin (3 x) + cos (3 x))) : - \frac{1}{3} x^{2} cos (3 x) + \frac{2}{9} x sin (3 x) - \frac{7}{27} cos (3 x)

= - \frac{1}{3} x^{2} cos (3 x) + \frac{2}{9} x sin (3 x) - \frac{7}{27} cos (3 x)

定数を解答に追加する

= - \frac{1}{3} x^{2} cos (3 x) + \frac{2}{9} x sin (3 x) - \frac{7}{27} cos (3 x) + C