integral of e^{i7x}cos(3x)

解

\int e^{i 7 x} cos (3 x) d x

\frac{1}{120} (- 14 sin (10 x) + 35 sin (4 x) + 40 cos (7 x) sin (3 x)) + i \frac{1}{120} (14 cos (10 x) - 35 cos (4 x) + 40 sin (3 x) sin (7 x)) + C

解答ステップ

\int e^{i 7 x} cos (3 x) d x

部分積分を適用する

= \frac{1}{3} sin (3 x) (cos (7 x) + i sin (7 x)) - \int \frac{1}{3} sin (3 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}{3} sin (3 x) (cos (7 x) + i sin (7 x)) - \frac{1}{3} \cdot \int sin (3 x) (- 7 sin (7 x) + 7 i cos (7 x)) d x

標準的な複素数形式で

sin (3 x) (- 7 sin (7 x) + 7 i cos (7 x))

を書き換える:

- 7 sin (3 x) sin (7 x) + 7 sin (3 x) cos (7 x) i

= \frac{1}{3} sin (3 x) (cos (7 x) + i sin (7 x)) - \frac{1}{3} \cdot \int - 7 sin (3 x) sin (7 x) + 7 i sin (3 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}{3} sin (3 x) (cos (7 x) + i sin (7 x)) - \frac{1}{3} (- \int 7 sin (3 x) sin (7 x) d x + \int 7 i sin (3 x) cos (7 x) d x)

\int 7 sin (3 x) sin (7 x) d x = \frac{7}{2} (\frac{1}{4} sin (4 x) - \frac{1}{10} sin (10 x))

\int 7 i sin (3 x) cos (7 x) d x = i \frac{1}{40} (- 14 cos (10 x) + 35 cos (4 x))

= \frac{1}{3} sin (3 x) (cos (7 x) + i sin (7 x)) - \frac{1}{3} (- \frac{7}{2} (\frac{1}{4} sin (4 x) - \frac{1}{10} sin (10 x)) + i \frac{1}{40} (- 14 cos (10 x) + 35 cos (4 x)))

簡素化

\frac{1}{3} sin (3 x) (cos (7 x) + i sin (7 x)) - \frac{1}{3} (- \frac{7}{2} (\frac{1}{4} sin (4 x) - \frac{1}{10} sin (10 x)) + i \frac{1}{40} (- 14 cos (10 x) + 35 cos (4 x))) : \frac{1}{120} (- 14 sin (10 x) + 35 sin (4 x) + 40 cos (7 x) sin (3 x)) + i \frac{1}{120} (14 cos (10 x) - 35 cos (4 x) + 40 sin (3 x) sin (7 x))

= \frac{1}{120} (- 14 sin (10 x) + 35 sin (4 x) + 40 cos (7 x) sin (3 x)) + i \frac{1}{120} (14 cos (10 x) - 35 cos (4 x) + 40 sin (3 x) sin (7 x))

定数を解答に追加する

= \frac{1}{120} (- 14 sin (10 x) + 35 sin (4 x) + 40 cos (7 x) sin (3 x)) + i \frac{1}{120} (14 cos (10 x) - 35 cos (4 x) + 40 sin (3 x) sin (7 x)) + C