What is the integral from 0 to t of e^{-2x}cos(3x) ?

The integral from 0 to t of e^{-2x}cos(3x) is (3e^{-2t}sin(3t)-2e^{-2t}cos(3t)+2)/(13)

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integral from 0 to t of e^{-2x}cos(3x)

\int_{0}^{t} e^{- 2 x} cos (3 x) d x

\frac{3 e ^{- 2 t} sin ( 3 t ) - 2 e ^{- 2 t} cos ( 3 t ) + 2}{13}

Solution steps

\int_{0}^{t} e^{- 2 x} cos (3 x) d x

Compute the indefinite integral:

\int e^{- 2 x} cos (3 x) d x = \frac{3 e ^{- 2 x} sin ( 3 x )}{13} - \frac{2 e ^{- 2 x} cos ( 3 x )}{13} + C

Compute the boundaries:

\int_{0}^{t} e^{- 2 x} cos (3 x) d x = \frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13} - (- \frac{2}{13})

= \frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13} - (- \frac{2}{13})

Simplify

= \frac{3 e ^{- 2 t} sin ( 3 t ) - 2 e ^{- 2 t} cos ( 3 t ) + 2}{13}

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What is the integral from 0 to t of e^{-2x}cos(3x) ?
The integral from 0 to t of e^{-2x}cos(3x) is (3e^{-2t}sin(3t)-2e^{-2t}cos(3t)+2)/(13)