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

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

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

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

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

Solution steps

\int sin (x) e^{- 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

\int 3 sin^{2} (x co) s^{3} x d x

\int_{0}^{4} \frac{1}{x + 2} d x

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 0

\int \frac{2 z}{1 - z ^{2}} d z

x \to 0 + lim (\frac{4}{x ^{2}})

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