integral of e^{i6x}cos(4x)

Lösung

\int e^{i 6 x} cos (4 x) d x

\frac{1}{40} (15 sin (2 x) - 3 sin (10 x) + 10 cos (6 x) sin (4 x)) + i \frac{1}{40} (- 15 cos (2 x) + 3 cos (10 x) + 10 sin (4 x) sin (6 x)) + C

Schritte zur Lösung

\int e^{i 6 x} cos (4 x) d x

Wende die partielle Integration an

= \frac{1}{4} sin (4 x) (cos (6 x) + i sin (6 x)) - \int \frac{3}{2} sin (4 x) (i cos (6 x) - sin (6 x)) d x

Entferne die Konstante:

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

= \frac{1}{4} sin (4 x) (cos (6 x) + i sin (6 x)) - \frac{3}{2} \cdot \int sin (4 x) (i cos (6 x) - sin (6 x)) d x

Schreibe

sin (4 x) (i cos (6 x) - sin (6 x))

in der Standard komplexen Form um:

- sin (4 x) sin (6 x) + sin (4 x) cos (6 x) i

= \frac{1}{4} sin (4 x) (cos (6 x) + i sin (6 x)) - \frac{3}{2} \cdot \int - sin (4 x) sin (6 x) + i sin (4 x) cos (6 x) d x

Wende die Summenregel an:

\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 (6 x) + i sin (6 x)) - \frac{3}{2} (- \int sin (4 x) sin (6 x) d x + \int i sin (4 x) cos (6 x) d x)

\int sin (4 x) sin (6 x) d x = \frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{10} sin (10 x))

\int i sin (4 x) cos (6 x) d x = i \frac{1}{20} (- cos (10 x) + 5 cos (2 x))

= \frac{1}{4} sin (4 x) (cos (6 x) + i sin (6 x)) - \frac{3}{2} (- \frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{10} sin (10 x)) + i \frac{1}{20} (- cos (10 x) + 5 cos (2 x)))

Vereinfache

\frac{1}{4} sin (4 x) (cos (6 x) + i sin (6 x)) - \frac{3}{2} (- \frac{1}{2} (\frac{1}{2} sin (2 x) - \frac{1}{10} sin (10 x)) + i \frac{1}{20} (- cos (10 x) + 5 cos (2 x))) : \frac{1}{40} (15 sin (2 x) - 3 sin (10 x) + 10 cos (6 x) sin (4 x)) + i \frac{1}{40} (- 15 cos (2 x) + 3 cos (10 x) + 10 sin (4 x) sin (6 x))

= \frac{1}{40} (15 sin (2 x) - 3 sin (10 x) + 10 cos (6 x) sin (4 x)) + i \frac{1}{40} (- 15 cos (2 x) + 3 cos (10 x) + 10 sin (4 x) sin (6 x))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{40} (15 sin (2 x) - 3 sin (10 x) + 10 cos (6 x) sin (4 x)) + i \frac{1}{40} (- 15 cos (2 x) + 3 cos (10 x) + 10 sin (4 x) sin (6 x)) + C

\int 1 + (- tan (x))^{2} d x

\frac{\partial}{\partial v} (\frac{v}{u})

d er i v a t i v e \frac{1}{x ^{17}}

\int_{- 2}^{0} \frac{x ^{2}}{\frac{16}{3}} d x

x \to 1 lim (\frac{x ^{2}}{1 - x ^{2}})