integral of e^{i2t}

Lösung

\int e^{i 2 t} d t

\frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t) + C

Schritte zur Lösung

\int e^{i 2 t} d t

Wende imaginäre Zahlenregel an:

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

= \int cos (2 t) + i sin (2 t) 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 (2 t) d t + \int i sin (2 t) d t

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

\int i sin (2 t) d t = - i \frac{1}{2} cos (2 t)

= \frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} sin (2 t) - i \frac{1}{2} cos (2 t) + C

\int tan^{4} (2 x) sec (2 x) d x

\int (\frac{6 + x ^{2}}{x}) d x

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\int \frac{( 6 x - 6 )}{x ^{2} - 2 x + 5} d x

\int (10 x + 2) d x