What is the integral from 0 to pi/4 of e^{-2it} ?

The integral from 0 to pi/4 of e^{-2it} is 1/2-i/2

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integral from 0 to pi/4 of e^{-2it}

\int_{0}^{\frac{π}{4}} e^{- 2 i t} d t

\frac{1}{2} - \frac{i}{2}

Solution steps

\int_{0}^{\frac{π}{4}} e^{- 2 i t} d t

Simplify

e^{- 2 i t} : cos (2 t) - i sin (2 t)

= \int_{0}^{\frac{π}{4}} cos (2 t) - i sin (2 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_{0}^{\frac{π}{4}} cos (2 t) d t - \int_{0}^{\frac{π}{4}} i sin (2 t) d t

\int_{0}^{\frac{π}{4}} cos (2 t) d t = \frac{1}{2}

\int_{0}^{\frac{π}{4}} i sin (2 t) d t = \frac{i}{2}

= \frac{1}{2} - \frac{i}{2}

\int_{0}^{\infty} e^{x} x d x

\int_{0}^{3} k x^{2} d x

\int_{1}^{3} \frac{2}{x ^{2} - 1} d x

\int_{1}^{\infty} n e^{- n^{2}}

What is the integral from 0 to pi/4 of e^{-2it} ?
The integral from 0 to pi/4 of e^{-2it} is 1/2-i/2