integral of e^{i7x}cos(5x)

解

\int e^{i 7 x} cos (5 x) d x

\frac{1}{120} (42 sin (2 x) - 7 sin (12 x) + 24 cos (7 x) sin (5 x)) + i \frac{1}{120} (- 42 cos (2 x) + 7 cos (12 x) + 24 sin (5 x) sin (7 x)) + C

解答ステップ

\int e^{i 7 x} cos (5 x) d x

部分積分を適用する

= \frac{1}{5} sin (5 x) (cos (7 x) + i sin (7 x)) - \int \frac{1}{5} sin (5 x) (- 7 sin (7 x) + 7 i cos (7 x)) d x

定数を除く:

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

= \frac{1}{5} sin (5 x) (cos (7 x) + i sin (7 x)) - \frac{1}{5} \cdot \int sin (5 x) (- 7 sin (7 x) + 7 i cos (7 x)) d x

標準的な複素数形式で

sin (5 x) (- 7 sin (7 x) + 7 i cos (7 x))

を書き換える:

- 7 sin (5 x) sin (7 x) + 7 sin (5 x) cos (7 x) i

= \frac{1}{5} sin (5 x) (cos (7 x) + i sin (7 x)) - \frac{1}{5} \cdot \int - 7 sin (5 x) sin (7 x) + 7 i sin (5 x) cos (7 x) d x

総和規則を適用する:

\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 (7 x) + i sin (7 x)) - \frac{1}{5} (- \int 7 sin (5 x) sin (7 x) d x + \int 7 i sin (5 x) cos (7 x) d x)

\int 7 sin (5 x) sin (7 x) d x = \frac{7}{2} (\frac{1}{2} sin (2 x) - \frac{1}{12} sin (12 x))

\int 7 i sin (5 x) cos (7 x) d x = i \frac{1}{24} (42 cos (2 x) - 7 cos (12 x))

= \frac{1}{5} sin (5 x) (cos (7 x) + i sin (7 x)) - \frac{1}{5} (- \frac{7}{2} (\frac{1}{2} sin (2 x) - \frac{1}{12} sin (12 x)) + i \frac{1}{24} (42 cos (2 x) - 7 cos (12 x)))

簡素化

\frac{1}{5} sin (5 x) (cos (7 x) + i sin (7 x)) - \frac{1}{5} (- \frac{7}{2} (\frac{1}{2} sin (2 x) - \frac{1}{12} sin (12 x)) + i \frac{1}{24} (42 cos (2 x) - 7 cos (12 x))) : \frac{1}{120} (42 sin (2 x) - 7 sin (12 x) + 24 cos (7 x) sin (5 x)) + i \frac{1}{120} (- 42 cos (2 x) + 7 cos (12 x) + 24 sin (5 x) sin (7 x))

= \frac{1}{120} (42 sin (2 x) - 7 sin (12 x) + 24 cos (7 x) sin (5 x)) + i \frac{1}{120} (- 42 cos (2 x) + 7 cos (12 x) + 24 sin (5 x) sin (7 x))

定数を解答に追加する

= \frac{1}{120} (42 sin (2 x) - 7 sin (12 x) + 24 cos (7 x) sin (5 x)) + i \frac{1}{120} (- 42 cos (2 x) + 7 cos (12 x) + 24 sin (5 x) sin (7 x)) + C