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

The integral of e^{-x/3}sin(2x) is -(18e^{-x/3}cos(2x))/(37)-(3e^{-x/3}sin(2x))/(37)+C

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

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

- \frac{18 e ^{- \frac{x}{3}} cos ( 2 x )}{37} - \frac{3 e ^{- \frac{x}{3}} sin ( 2 x )}{37} + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= - \frac{18 e ^{- \frac{x}{3}} cos ( 2 x )}{37} - \frac{3 e ^{- \frac{x}{3}} sin ( 2 x )}{37}

Add a constant to the solution

= - \frac{18 e ^{- \frac{x}{3}} cos ( 2 x )}{37} - \frac{3 e ^{- \frac{x}{3}} sin ( 2 x )}{37} + C

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What is the integral of e^{-x/3}sin(2x) ?
The integral of e^{-x/3}sin(2x) is -(18e^{-x/3}cos(2x))/(37)-(3e^{-x/3}sin(2x))/(37)+C