What is the integral of e^{-6x}sin(3x) ?

The integral of e^{-6x}sin(3x) is -(e^{-6x}cos(3x))/(15)-(2e^{-6x}sin(3x))/(15)+C

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integral of e^{-6x}sin(3x)

\int e^{- 6 x} sin (3 x) d x

- \frac{e ^{- 6 x} cos ( 3 x )}{15} - \frac{2 e ^{- 6 x} sin ( 3 x )}{15} + C

Solution steps

\int e^{- 6 x} sin (3 x) d x

Apply Integration By Parts

: - \frac{1}{3} e^{- 6 x} cos (3 x) - \int 2 e^{- 6 x} cos (3 x) d x

= - \frac{1}{3} e^{- 6 x} cos (3 x) - \int 2 e^{- 6 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^{- 6 x} cos (3 x) - 2 \cdot \int e^{- 6 x} cos (3 x) d x

Apply Integration By Parts

: \frac{1}{3} e^{- 6 x} sin (3 x) - \int - 2 e^{- 6 x} sin (3 x) d x

= - \frac{1}{3} e^{- 6 x} cos (3 x) - 2 (\frac{1}{3} e^{- 6 x} sin (3 x) - \int - 2 e^{- 6 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^{- 6 x} cos (3 x) - 2 (\frac{1}{3} e^{- 6 x} sin (3 x) - (- 2 \cdot \int e^{- 6 x} sin (3 x) d x))

Therefore

\int e^{- 6 x} sin (3 x) d x = - \frac{1}{3} e^{- 6 x} cos (3 x) - 2 (\frac{1}{3} e^{- 6 x} sin (3 x) - (- 2 \cdot \int e^{- 6 x} sin (3 x) d x))

Isolate

\int e^{- 6 x} sin (3 x) d x

= - \frac{e ^{- 6 x} cos ( 3 x )}{15} - \frac{2 e ^{- 6 x} sin ( 3 x )}{15}

Add a constant to the solution

= - \frac{e ^{- 6 x} cos ( 3 x )}{15} - \frac{2 e ^{- 6 x} sin ( 3 x )}{15} + C

x \to 3 lim (x^{2} + 4 x - 6)

p (x) = x e^{x, x}

\int sin^{5} (3 x) d x

\frac{d y}{d t} - \frac{4}{t} \cdot y = t^{5}

\int \frac{cos ( 3 π ) x}{2} d x

What is the integral of e^{-6x}sin(3x) ?
The integral of e^{-6x}sin(3x) is -(e^{-6x}cos(3x))/(15)-(2e^{-6x}sin(3x))/(15)+C