What is the integral of e^{-2x}sin(6x) ?

The integral of e^{-2x}sin(6x) is -(3e^{-2x}cos(6x))/(20)-(e^{-2x}sin(6x))/(20)+C

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integral of e^{-2x}sin(6x)

\int e^{- 2 x} sin (6 x) d x

- \frac{3 e ^{- 2 x} cos ( 6 x )}{20} - \frac{e ^{- 2 x} sin ( 6 x )}{20} + C

Solution steps

\int e^{- 2 x} sin (6 x) d x

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{- 2 x} sin (6 x) d x

= - \frac{3 e ^{- 2 x} cos ( 6 x )}{20} - \frac{e ^{- 2 x} sin ( 6 x )}{20}

Add a constant to the solution

= - \frac{3 e ^{- 2 x} cos ( 6 x )}{20} - \frac{e ^{- 2 x} sin ( 6 x )}{20} + C

\frac{d}{d x} (x^{6} e^{2.5 x})

\frac{d}{d x} (e^{- 0.02 x})

18 e^{x^{2}} x

\int \frac{e ^{x}}{2 x} d x

y^{^{''}} + 2 y^{^{'}} + 4 y = 5 sin (t)

What is the integral of e^{-2x}sin(6x) ?
The integral of e^{-2x}sin(6x) is -(3e^{-2x}cos(6x))/(20)-(e^{-2x}sin(6x))/(20)+C