integral of x^3sin(6x)

解答

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

- \frac{1}{6} x^{3} cos (6 x) + \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

求解步骤

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

使用分布积分法

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

\int - \frac{1}{2} x^{2} cos (6 x) d x = - \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

= - \frac{1}{6} x^{3} cos (6 x) - (- \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)))

化简

= - \frac{1}{6} x^{3} cos (6 x) + \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x))

解答补常数

= - \frac{1}{6} x^{3} cos (6 x) + \frac{1}{216} (18 x^{2} sin (6 x) - sin (6 x) + 6 x cos (6 x)) + C

t an g e n t f (x) = x^{2} - \frac{1}{x}

\frac{d}{d x} (1 - \frac{1}{1 + x})

\frac{\partial}{\partial x} (\frac{4 x - 4 z}{3 y + 4 z})

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

in v erse l a pl a ce \frac{3 s}{( s - 1 ) ( s ^{2} - 4 )}