What is the integral from 0 to pi/2 of e^{-t}sin(t) ?

The integral from 0 to pi/2 of e^{-t}sin(t) is -1/(2e^{pi/2)}+1/2

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integral from 0 to pi/2 of e^{-t}sin(t)

\int_{0}^{\frac{π}{2}} e^{- t} sin (t) d t

- \frac{1}{2 e ^{\frac{π}{2}}} + \frac{1}{2}

Decimal Notation

0.39606 \dots

Solution steps

\int_{0}^{\frac{π}{2}} e^{- t} sin (t) d t

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{\frac{π}{2}} e^{- t} sin (t) d t = - \frac{1}{2 e ^{\frac{π}{2}}} - (- \frac{1}{2})

= - \frac{1}{2 e ^{\frac{π}{2}}} - (- \frac{1}{2})

Simplify

= - \frac{1}{2 e ^{\frac{π}{2}}} + \frac{1}{2}

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What is the integral from 0 to pi/2 of e^{-t}sin(t) ?
The integral from 0 to pi/2 of e^{-t}sin(t) is -1/(2e^{pi/2)}+1/2