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

The integral of e^{-2x}cos(3x) is (3e^{-2x}sin(3x))/(13)-(2e^{-2x}cos(3x))/(13)+C

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integral of e^{-2x}cos(3x)

\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

Solution steps

\int e^{- 2 x} cos (3 x) d x

Apply Integration By Parts

: \frac{1}{3} e^{- 2 x} sin (3 x) - \int - \frac{2}{3} e^{- 2 x} sin (3 x) d x

= \frac{1}{3} e^{- 2 x} sin (3 x) - \int - \frac{2}{3} e^{- 2 x} sin (3 x) d x

Take the constant out:

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

= \frac{1}{3} e^{- 2 x} sin (3 x) - (- \frac{2}{3} \cdot \int e^{- 2 x} sin (3 x) d x)

Apply Integration By Parts

: - \frac{1}{3} e^{- 2 x} cos (3 x) - \int \frac{2}{3} e^{- 2 x} cos (3 x) d x

= \frac{1}{3} e^{- 2 x} sin (3 x) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 x} cos (3 x) - \int \frac{2}{3} e^{- 2 x} cos (3 x) d x))

Take the constant out:

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

= \frac{1}{3} e^{- 2 x} sin (3 x) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 x} cos (3 x) - \frac{2}{3} \cdot \int e^{- 2 x} cos (3 x) d x))

Therefore

\int e^{- 2 x} cos (3 x) d x = \frac{1}{3} e^{- 2 x} sin (3 x) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 x} cos (3 x) - \frac{2}{3} \cdot \int e^{- 2 x} cos (3 x) d x))

Isolate

\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}

Add a constant to the solution

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

x \to 3 lim (x^{2} + 16)

\frac{\partial}{\partial x} (25 - 5 x^{2} - y^{2})

x \to \infty lim (2 x + 5)

\int_{0}^{π} 3 sin^{2} (t co) s^{4} t d t

f (x) = \frac{3 x ^{2} e ^{x} - 4}{x ^{2}}

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