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

The integral of e^{-x}sin(pix) is -(pie^{-x}cos(pix))/(pi^2+1)-(e^{-x}sin(pix))/(pi^2+1)+C

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

\int e^{- x} sin (π x) d x

- \frac{π e ^{- x} cos ( π x )}{π ^{2} + 1} - \frac{e ^{- x} sin ( π x )}{π ^{2} + 1} + C

Solution steps

\int e^{- x} sin (π x) d x

Apply Integration By Parts

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

= - \frac{1}{π} e^{- x} cos (π x) - \int \frac{1}{π} e^{- x} cos (π x) d x

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{- x} sin (π x) d x

= - \frac{π e ^{- x} cos ( π x )}{π ^{2} + 1} - \frac{e ^{- x} sin ( π x )}{π ^{2} + 1}

Add a constant to the solution

= - \frac{π e ^{- x} cos ( π x )}{π ^{2} + 1} - \frac{e ^{- x} sin ( π x )}{π ^{2} + 1} + C

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What is the integral of e^{-x}sin(pix) ?
The integral of e^{-x}sin(pix) is -(pie^{-x}cos(pix))/(pi^2+1)-(e^{-x}sin(pix))/(pi^2+1)+C