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

The integral of e^{-x}sin(2x) is -(2e^{-x}cos(2x))/5-(e^{-x}sin(2x))/5+C

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

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

- \frac{2 e ^{- x} cos ( 2 x )}{5} - \frac{e ^{- x} sin ( 2 x )}{5} + C

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

= - \frac{2 e ^{- x} cos ( 2 x )}{5} - \frac{e ^{- x} sin ( 2 x )}{5}

Add a constant to the solution

= - \frac{2 e ^{- x} cos ( 2 x )}{5} - \frac{e ^{- x} sin ( 2 x )}{5} + C

\frac{\partial}{\partial x} (e^{x y} tan (y))

h \to 0 lim (\frac{( 18 h + 1 - 1 )}{h})

\int \frac{9 x ^{2} + 16}{x ^{4}} d x

\int 2 x (x^{2} - 1) d x

y = (\frac{x ^{2} + 3}{x ^{2} - 3})^{7}

What is the integral of e^{-x}sin(2x) ?
The integral of e^{-x}sin(2x) is -(2e^{-x}cos(2x))/5-(e^{-x}sin(2x))/5+C