What is the integral of e^{-2x}cos(5x) ?

The integral of e^{-2x}cos(5x) is (5e^{-2x}sin(5x))/(29)-(2e^{-2x}cos(5x))/(29)+C

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integral of e^{-2x}cos(5x)

\int e^{- 2 x} cos (5 x) d x

\frac{5 e ^{- 2 x} sin ( 5 x )}{29} - \frac{2 e ^{- 2 x} cos ( 5 x )}{29} + C

Solution steps

\int e^{- 2 x} cos (5 x) d x

Apply Integration By Parts

: \frac{1}{5} e^{- 2 x} sin (5 x) - \int - \frac{2}{5} e^{- 2 x} sin (5 x) d x

= \frac{1}{5} e^{- 2 x} sin (5 x) - \int - \frac{2}{5} e^{- 2 x} sin (5 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} e^{- 2 x} sin (5 x) - (- \frac{2}{5} \cdot \int e^{- 2 x} sin (5 x) d x)

Apply Integration By Parts

: - \frac{1}{5} e^{- 2 x} cos (5 x) - \int \frac{2}{5} e^{- 2 x} cos (5 x) d x

= \frac{1}{5} e^{- 2 x} sin (5 x) - (- \frac{2}{5} (- \frac{1}{5} e^{- 2 x} cos (5 x) - \int \frac{2}{5} e^{- 2 x} cos (5 x) d x))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} e^{- 2 x} sin (5 x) - (- \frac{2}{5} (- \frac{1}{5} e^{- 2 x} cos (5 x) - \frac{2}{5} \cdot \int e^{- 2 x} cos (5 x) d x))

Therefore

\int e^{- 2 x} cos (5 x) d x = \frac{1}{5} e^{- 2 x} sin (5 x) - (- \frac{2}{5} (- \frac{1}{5} e^{- 2 x} cos (5 x) - \frac{2}{5} \cdot \int e^{- 2 x} cos (5 x) d x))

Isolate

\int e^{- 2 x} cos (5 x) d x

= \frac{5 e ^{- 2 x} sin ( 5 x )}{29} - \frac{2 e ^{- 2 x} cos ( 5 x )}{29}

Add a constant to the solution

= \frac{5 e ^{- 2 x} sin ( 5 x )}{29} - \frac{2 e ^{- 2 x} cos ( 5 x )}{29} + C

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What is the integral of e^{-2x}cos(5x) ?
The integral of e^{-2x}cos(5x) is (5e^{-2x}sin(5x))/(29)-(2e^{-2x}cos(5x))/(29)+C