integral of 4xcos(2-3x)

解答

\int 4 x cos (2 - 3 x) d x

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

求解步骤

\int 4 x cos (2 - 3 x) d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x cos (2 - 3 x) d x

使用分布积分法

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

\int - \frac{1}{3} sin (2 - 3 x) d x = - \frac{1}{9} cos (2 - 3 x)

= 4 (- \frac{1}{3} x sin (2 - 3 x) - (- \frac{1}{9} cos (2 - 3 x)))

化简

= 4 (- \frac{1}{3} x sin (2 - 3 x) + \frac{1}{9} cos (2 - 3 x))

解答补常数

= 4 (- \frac{1}{3} x sin (2 - 3 x) + \frac{1}{9} cos (2 - 3 x)) + C

\int (8 x - 5) d x

(cos (2 t))^{^{'}}

\int (y^{\frac{5}{2}} - 5 y^{\frac{4}{3}} - 2 y^{\frac{1}{4}} - y) d y

y^{^{'}} = y (y + 1)

in v erse l a pl a ce \frac{2}{s ^{2} + 9}