integral of x^4cos(2x)

解答

\int x^{4} cos (2 x) d x

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

求解步骤

\int x^{4} cos (2 x) d x

使用分布积分法

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

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

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

化简

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

解答补常数

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

a re a 9 - x^{2}, y = 0

\frac{d}{d x} (x sin (6 x))

\int 16 y^{3} + 9 y^{2} - 6 y + 3 d y

d er i v a t i v e f (x) = x^{4} 7 - 9 x

\int_{0}^{1} 4^{4 x} d x