integral of e^{iwt+ibt}

Lösung

\int e^{i wt + ib t} d t

\frac{1}{b + w} sin (b t + wt) - i \frac{1}{b + w} cos (b t + wt) + C

Schritte zur Lösung

\int e^{i wt + ib t} d t

Wende imaginäre Zahlenregel an:

e^{ia} = cos (a) + i sin (a)

= \int cos (b t + wt) + i sin (b t + wt) d t

Wende die Summenregel an:

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

= \int cos (b t + wt) d t + \int i sin (b t + wt) d t

\int cos (b t + wt) d t = \frac{1}{b + w} sin (b t + wt)

\int i sin (b t + wt) d t = - i \frac{1}{b + w} cos (b t + wt)

= \frac{1}{b + w} sin (b t + wt) - i \frac{1}{b + w} cos (b t + wt)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{b + w} sin (b t + wt) - i \frac{1}{b + w} cos (b t + wt) + C

x \to 3 lim (x + 4)

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

x \to 5 + lim (\frac{- 2}{x ^{2} - 25})

\int x^{2} (x + 3) d x

s l o p e x^{2} + y^{2} = 25, (4, 3)