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

The integral from 0 to pi of e^{-x}cos(2x) is -1/(5e^pi)+1/5

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

\int_{0}^{π} e^{- x} cos (2 x) d x

- \frac{1}{5 e ^{π}} + \frac{1}{5}

Solution steps

\int_{0}^{π} e^{- x} cos (2 x) d x

Compute the indefinite integral:

\int e^{- x} cos (2 x) d x = \frac{2 e ^{- x} sin ( 2 x )}{5} - \frac{e ^{- x} cos ( 2 x )}{5} + C

Compute the boundaries:

\int_{0}^{π} e^{- x} cos (2 x) d x = - \frac{1}{5 e ^{π}} - (- \frac{1}{5})

= - \frac{1}{5 e ^{π}} - (- \frac{1}{5})

Simplify

= - \frac{1}{5 e ^{π}} + \frac{1}{5}

\frac{\partial}{\partial y} (ln (x + e^{- y}))

\int cos (a ln (\frac{x}{b})) d x

f (x) = \frac{1}{x - 2}

y^{^{'}} - 2 x (y^{2} + 1) = 0

What is the integral from 0 to pi of e^{-x}cos(2x) ?
The integral from 0 to pi of e^{-x}cos(2x) is -1/(5e^pi)+1/5