What is the integral of e^{(1+i)t} ?

The integral of e^{(1+i)t} is 1/2 (e^tcos(t)+e^tsin(t))+i 1/2 (-e^tcos(t)+e^tsin(t))+C

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integral of e^{(1+i)t}

\int e^{(1 + i) t} d t

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

Solution steps

\int e^{(1 + i) t} d t

Simplify

e^{(1 + i) t} : e^{t} cos (t) + i e^{t} sin (t)

= \int e^{t} cos (t) + i e^{t} sin (t) d t

Apply the Sum Rule:

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

= \int e^{t} cos (t) d t + \int i e^{t} sin (t) d t

\int e^{t} cos (t) d t = \frac{e ^{t} sin ( t )}{2} + \frac{e ^{t} cos ( t )}{2}

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

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

Simplify

\frac{e ^{t} sin ( t )}{2} + \frac{e ^{t} cos ( t )}{2} + i \frac{1}{2} (- e^{t} cos (t) + e^{t} sin (t)) : \frac{1}{2} (e^{t} cos (t) + e^{t} sin (t)) + i \frac{1}{2} (- e^{t} cos (t) + e^{t} sin (t))

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

Add a constant to the solution

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

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\int x^{4} cos (5 x) d x

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What is the integral of e^{(1+i)t} ?
The integral of e^{(1+i)t} is 1/2 (e^tcos(t)+e^tsin(t))+i 1/2 (-e^tcos(t)+e^tsin(t))+C