What is the integral of (sin(2x))/(e^{-4x)} ?

The integral of (sin(2x))/(e^{-4x)} is -(e^{4x}cos(2x)}{10}+\frac{e^{4x}sin(2x))/5+C

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integral of (sin(2x))/(e^{-4x)}

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

- \frac{e ^{4 x} cos ( 2 x )}{10} + \frac{e ^{4 x} sin ( 2 x )}{5} + C

Solution steps

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

Simplify

\frac{sin ( 2 x )}{e ^{- 4 x}} : e^{4 x} sin (2 x)

= \int e^{4 x} sin (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} e^{4 x} cos (2 x) - \int - 2 e^{4 x} cos (2 x) d x

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

\int e^{4 x} sin (2 x) d x

= - \frac{e ^{4 x} cos ( 2 x )}{10} + \frac{e ^{4 x} sin ( 2 x )}{5}

Add a constant to the solution

= - \frac{e ^{4 x} cos ( 2 x )}{10} + \frac{e ^{4 x} sin ( 2 x )}{5} + C

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What is the integral of (sin(2x))/(e^{-4x)} ?
The integral of (sin(2x))/(e^{-4x)} is -(e^{4x}cos(2x)}{10}+\frac{e^{4x}sin(2x))/5+C