What is the integral of (sin(2x))/(e^x) ?

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

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integral of (sin(2x))/(e^x)

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int sin (2 x) e^{- 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{d}{d x} (\frac{4 e ^{x}}{2 e ^{x} + 1})

2 sin (2 t)

\frac{d y}{d x} y = \frac{x ^{3}}{3} - x

\frac{\partial}{\partial x} (cos^{6} (x^{8} y^{9}))

y^{^{''}} + 5 y^{^{'}} + 4 y = 0

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