integral of x^3cos(6x)

解答

\int x^{3} cos (6 x) d x

\frac{1}{6} x^{3} sin (6 x) - \frac{1}{2} (- \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x))) + C

求解步骤

\int x^{3} cos (6 x) d x

使用分布积分法

= \frac{1}{6} x^{3} sin (6 x) - \int \frac{1}{2} x^{2} sin (6 x) d x

\int \frac{1}{2} x^{2} sin (6 x) d x = \frac{1}{2} (- \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x)))

= \frac{1}{6} x^{3} sin (6 x) - \frac{1}{2} (- \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x)))

解答补常数

= \frac{1}{6} x^{3} sin (6 x) - \frac{1}{2} (- \frac{1}{6} x^{2} cos (6 x) + \frac{1}{108} (6 x sin (6 x) + cos (6 x))) + C

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

d er i v a t i v e f (x) = - 2 x sin (x^{2})

\frac{\partial}{\partial c} (\frac{a ^{3} b ^{4}}{c ^{2} d ^{4}})

f (x) = arccos (x)

\frac{d}{d x} (3 x e^{- 3 x})