What is the integral of e^{-t}cos(pit) ?

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

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integral of e^{-t}cos(pit)

\int e^{- t} cos (π t) d t

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

Solution steps

\int e^{- t} cos (π t) d t

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{- t} cos (π t) d t

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

Add a constant to the solution

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

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