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

解答

\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

求解步骤

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

化简

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

使用积分加法定则:

\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))

化简

\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))

解答补常数

= \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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