What is the integral from 0 to pi/2 of e^{-9x}cos(x) ?

The integral from 0 to pi/2 of e^{-9x}cos(x) is 1/(82e^{\frac{9pi){2}}}+9/82

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integral from 0 to pi/2 of e^{-9x}cos(x)

\int_{0}^{\frac{π}{2}} e^{- 9 x} cos (x) d x

\frac{1}{82 e ^{\frac{9 π}{2}}} + \frac{9}{82}

Decimal Notation

0.10975 \dots

Solution steps

\int_{0}^{\frac{π}{2}} e^{- 9 x} cos (x) d x

Compute the indefinite integral:

\int e^{- 9 x} cos (x) d x = \frac{e ^{- 9 x} sin ( x )}{82} - \frac{9 e ^{- 9 x} cos ( x )}{82} + C

Compute the boundaries:

\int_{0}^{\frac{π}{2}} e^{- 9 x} cos (x) d x = \frac{1}{82 e ^{\frac{9 π}{2}}} - (- \frac{9}{82})

= \frac{1}{82 e ^{\frac{9 π}{2}}} - (- \frac{9}{82})

Simplify

= \frac{1}{82 e ^{\frac{9 π}{2}}} + \frac{9}{82}

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What is the integral from 0 to pi/2 of e^{-9x}cos(x) ?
The integral from 0 to pi/2 of e^{-9x}cos(x) is 1/(82e^{\frac{9pi){2}}}+9/82