What is the integral of e^{-3x}cos(4x) ?

The integral of e^{-3x}cos(4x) is (4e^{-3x}sin(4x))/(25)-(3e^{-3x}cos(4x))/(25)+C

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integral of e^{-3x}cos(4x)

\int e^{- 3 x} cos (4 x) d x

\frac{4 e ^{- 3 x} sin ( 4 x )}{25} - \frac{3 e ^{- 3 x} cos ( 4 x )}{25} + C

Solution steps

\int e^{- 3 x} cos (4 x) d x

Apply Integration By Parts

: \frac{1}{4} e^{- 3 x} sin (4 x) - \int - \frac{3}{4} e^{- 3 x} sin (4 x) d x

= \frac{1}{4} e^{- 3 x} sin (4 x) - \int - \frac{3}{4} e^{- 3 x} sin (4 x) d x

Take the constant out:

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

= \frac{1}{4} e^{- 3 x} sin (4 x) - (- \frac{3}{4} \cdot \int e^{- 3 x} sin (4 x) d x)

Apply Integration By Parts

: - \frac{1}{4} e^{- 3 x} cos (4 x) - \int \frac{3}{4} e^{- 3 x} cos (4 x) d x

= \frac{1}{4} e^{- 3 x} sin (4 x) - (- \frac{3}{4} (- \frac{1}{4} e^{- 3 x} cos (4 x) - \int \frac{3}{4} e^{- 3 x} cos (4 x) d x))

Take the constant out:

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

= \frac{1}{4} e^{- 3 x} sin (4 x) - (- \frac{3}{4} (- \frac{1}{4} e^{- 3 x} cos (4 x) - \frac{3}{4} \cdot \int e^{- 3 x} cos (4 x) d x))

Therefore

\int e^{- 3 x} cos (4 x) d x = \frac{1}{4} e^{- 3 x} sin (4 x) - (- \frac{3}{4} (- \frac{1}{4} e^{- 3 x} cos (4 x) - \frac{3}{4} \cdot \int e^{- 3 x} cos (4 x) d x))

Isolate

\int e^{- 3 x} cos (4 x) d x

= \frac{4 e ^{- 3 x} sin ( 4 x )}{25} - \frac{3 e ^{- 3 x} cos ( 4 x )}{25}

Add a constant to the solution

= \frac{4 e ^{- 3 x} sin ( 4 x )}{25} - \frac{3 e ^{- 3 x} cos ( 4 x )}{25} + C

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What is the integral of e^{-3x}cos(4x) ?
The integral of e^{-3x}cos(4x) is (4e^{-3x}sin(4x))/(25)-(3e^{-3x}cos(4x))/(25)+C