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

The integral of e^{-x}cos(npix) is (pine^{-x}sin(pinx))/(pi^2n^2+1)-(e^{-x}cos(pinx))/(pi^2n^2+1)+C

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integral of e^{-x}cos(npix)

\int e^{- x} cos (nπ x) d x

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

Solution steps

\int e^{- x} cos (nπ x) d x

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{- x} cos (πn x) d x

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

Add a constant to the solution

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

What is the integral of e^{-x}cos(npix) ?
The integral of e^{-x}cos(npix) is (pine^{-x}sin(pinx))/(pi^2n^2+1)-(e^{-x}cos(pinx))/(pi^2n^2+1)+C