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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\frac{d}{d x} (csc^{2} (x) cot (x))

\frac{d}{d x} (7 x (x^{3} - 4 x + 5))

\int \frac{u ^{3}}{1 + u ^{2}} d u

\int π x d x

t \to + 0 lim (\frac{2 t}{tan ( t )})

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