integral of e^{i6x}cos(5x)

解答

\int e^{i 6 x} cos (5 x) d x

\frac{1}{55} (11 (3 sin (x) + cos (6 x) sin (5 x)) - 3 sin (11 x)) + i \frac{1}{55} (11 (- 3 cos (x) + sin (5 x) sin (6 x)) + 3 cos (11 x)) + C

求解步骤

\int e^{i 6 x} cos (5 x) d x

使用分布积分法

= \frac{1}{5} sin (5 x) (cos (6 x) + i sin (6 x)) - \int \frac{1}{5} sin (5 x) (- 6 sin (6 x) + 6 i cos (6 x)) d x

提出常数:

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

= \frac{1}{5} sin (5 x) (cos (6 x) + i sin (6 x)) - \frac{1}{5} \cdot \int sin (5 x) (- 6 sin (6 x) + 6 i cos (6 x)) d x

将

sin (5 x) (- 6 sin (6 x) + 6 i cos (6 x))

改写成标准复数形式:

- 6 sin (5 x) sin (6 x) + 6 sin (5 x) cos (6 x) i

= \frac{1}{5} sin (5 x) (cos (6 x) + i sin (6 x)) - \frac{1}{5} \cdot \int - 6 sin (5 x) sin (6 x) + 6 i sin (5 x) cos (6 x) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{5} sin (5 x) (cos (6 x) + i sin (6 x)) - \frac{1}{5} (- \int 6 sin (5 x) sin (6 x) d x + \int 6 i sin (5 x) cos (6 x) d x)

\int 6 sin (5 x) sin (6 x) d x = 3 (sin (x) - \frac{1}{11} sin (11 x))

\int 6 i sin (5 x) cos (6 x) d x = i \frac{1}{11} (33 cos (x) - 3 cos (11 x))

= \frac{1}{5} sin (5 x) (cos (6 x) + i sin (6 x)) - \frac{1}{5} (- 3 (sin (x) - \frac{1}{11} sin (11 x)) + i \frac{1}{11} (33 cos (x) - 3 cos (11 x)))

化简

\frac{1}{5} sin (5 x) (cos (6 x) + i sin (6 x)) - \frac{1}{5} (- 3 (sin (x) - \frac{1}{11} sin (11 x)) + i \frac{1}{11} (33 cos (x) - 3 cos (11 x))) : \frac{1}{55} (11 (3 sin (x) + cos (6 x) sin (5 x)) - 3 sin (11 x)) + i \frac{1}{55} (11 (- 3 cos (x) + sin (5 x) sin (6 x)) + 3 cos (11 x))

= \frac{1}{55} (11 (3 sin (x) + cos (6 x) sin (5 x)) - 3 sin (11 x)) + i \frac{1}{55} (11 (- 3 cos (x) + sin (5 x) sin (6 x)) + 3 cos (11 x))

解答补常数

= \frac{1}{55} (11 (3 sin (x) + cos (6 x) sin (5 x)) - 3 sin (11 x)) + i \frac{1}{55} (11 (- 3 cos (x) + sin (5 x) sin (6 x)) + 3 cos (11 x)) + C

\int x^{2} (8 x - 3) d x

\int \frac{x ^{3} + 2}{x ^{3}} x - \frac{1}{x ^{2}} d x

\int sin^{6} (5 x) d x

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

n = 1 \sum \infty (\frac{2}{3})^{n + 1}