What is the integral from 0 to 1 of e^{-x}cos(wx) ?

The integral from 0 to 1 of e^{-x}cos(wx) is (wsin(w)-cos(w))/(e(w^2+1))+1/(w^2+1)

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integral from 0 to 1 of e^{-x}cos(wx)

\int_{0}^{1} e^{- x} cos (w x) d x

\frac{w sin ( w ) - cos ( w )}{e ( w ^{2} + 1 )} + \frac{1}{w ^{2} + 1}

Solution steps

\int_{0}^{1} e^{- x} cos (w x) d x

Compute the indefinite integral:

\int e^{- x} cos (w x) d x = \frac{w e ^{- x} sin ( w x )}{w ^{2} + 1} - \frac{e ^{- x} cos ( w x )}{w ^{2} + 1} + C

Compute the boundaries:

\int_{0}^{1} e^{- x} cos (w x) d x = \frac{w sin ( w )}{e ( w ^{2} + 1 )} - \frac{cos ( w )}{e ( w ^{2} + 1 )} - (- \frac{1}{w ^{2} + 1})

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

Simplify

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

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\int_{- 1}^{0} x sec^{2} (x) d x

\int_{0}^{2} (\frac{x ^{2}}{2} + 2) d x

\int_{0}^{3} 0.5 e^{- 0.5 x} d x

What is the integral from 0 to 1 of e^{-x}cos(wx) ?
The integral from 0 to 1 of e^{-x}cos(wx) is (wsin(w)-cos(w))/(e(w^2+1))+1/(w^2+1)