integral of e^{i8x}cos(4x)

解答

\int e^{i 8 x} cos (4 x) d x

\frac{1}{12} (- sin (12 x) + 3 (sin (4 x) + cos (8 x) sin (4 x))) + i \frac{1}{12} (cos (12 x) + 3 (- cos (4 x) + sin (4 x) sin (8 x))) + C

求解步骤

\int e^{i 8 x} cos (4 x) d x

使用分布积分法

= \frac{1}{4} sin (4 x) (cos (8 x) + i sin (8 x)) - \int 2 sin (4 x) (i cos (8 x) - sin (8 x)) d x

提出常数:

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

= \frac{1}{4} sin (4 x) (cos (8 x) + i sin (8 x)) - 2 \cdot \int sin (4 x) (i cos (8 x) - sin (8 x)) d x

将

sin (4 x) (i cos (8 x) - sin (8 x))

改写成标准复数形式:

- sin (4 x) sin (8 x) + sin (4 x) cos (8 x) i

= \frac{1}{4} sin (4 x) (cos (8 x) + i sin (8 x)) - 2 \cdot \int - sin (4 x) sin (8 x) + i sin (4 x) cos (8 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}{4} sin (4 x) (cos (8 x) + i sin (8 x)) - 2 (- \int sin (4 x) sin (8 x) d x + \int i sin (4 x) cos (8 x) d x)

\int sin (4 x) sin (8 x) d x = \frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{12} sin (12 x))

\int i sin (4 x) cos (8 x) d x = i \frac{1}{24} (- cos (12 x) + 3 cos (4 x))

= \frac{1}{4} sin (4 x) (cos (8 x) + i sin (8 x)) - 2 (- \frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{12} sin (12 x)) + i \frac{1}{24} (- cos (12 x) + 3 cos (4 x)))

化简

\frac{1}{4} sin (4 x) (cos (8 x) + i sin (8 x)) - 2 (- \frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{12} sin (12 x)) + i \frac{1}{24} (- cos (12 x) + 3 cos (4 x))) : \frac{1}{12} (- sin (12 x) + 3 (sin (4 x) + cos (8 x) sin (4 x))) + i \frac{1}{12} (cos (12 x) + 3 (- cos (4 x) + sin (4 x) sin (8 x)))

= \frac{1}{12} (- sin (12 x) + 3 (sin (4 x) + cos (8 x) sin (4 x))) + i \frac{1}{12} (cos (12 x) + 3 (- cos (4 x) + sin (4 x) sin (8 x)))

解答补常数

= \frac{1}{12} (- sin (12 x) + 3 (sin (4 x) + cos (8 x) sin (4 x))) + i \frac{1}{12} (cos (12 x) + 3 (- cos (4 x) + sin (4 x) sin (8 x))) + C

d er i v a t i v e f (x) = \frac{x ^{2} - 5 x}{x + 1}

\int 3 x^{2} x^{3} - 5 d x

\frac{d}{d x} (\frac{- x}{x ^{2} + y ^{2}})

\int \frac{6 x - 15}{x ^{2} - 3 x} d x

\int (2 sin (x) + x) d x