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

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

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

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

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

Solution steps

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

f (x) = x^{2} sin (2 x)

\int \frac{1}{- 2 x ^{2} - x + 1} d x

y^{^{''}} - y^{^{'}} = 1

\int \frac{4 ^{x} + 1 0 ^{x}}{2 ^{x}} d x

\int \frac{1}{x ( 1 + x )} d x

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