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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{5 n}{4 n + 15}

x \to - 1 - lim (\frac{x + 1}{x ^{4} - 1})

n = 0 \sum \infty \frac{1}{2 ^{n - 1}}

\frac{x ^{4}}{2} - \frac{2}{x ^{4}} + 6 x

\frac{d}{d x} (arcsin (5 x - 1))

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