What is the integral from-1 to 1 of e^{-x}cos(npix) ?

The integral from-1 to 1 of e^{-x}cos(npix) is -((-1)^n)/(e(pi^2n^2+1))+(e(-1)^n)/(pi^2n^2+1)

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integral from-1 to 1 of e^{-x}cos(npix)

\int_{- 1}^{1} e^{- x} cos (nπ x) d x

- \frac{( - 1 ) ^{n}}{e ( π ^{2} n ^{2} + 1 )} + \frac{e ( - 1 ) ^{n}}{π ^{2} n ^{2} + 1}

Solution steps

\int_{- 1}^{1} e^{- x} cos (nπ x) d x

Compute the indefinite integral:

\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

Compute the boundaries:

\int_{- 1}^{1} e^{- x} cos (nπ x) d x = - \frac{( - 1 ) ^{n}}{e ( π ^{2} n ^{2} + 1 )} - (- \frac{e ( - 1 ) ^{n}}{π ^{2} n ^{2} + 1})

= - \frac{( - 1 ) ^{n}}{e ( π ^{2} n ^{2} + 1 )} - (- \frac{e ( - 1 ) ^{n}}{π ^{2} n ^{2} + 1})

Simplify

= - \frac{( - 1 ) ^{n}}{e ( π ^{2} n ^{2} + 1 )} + \frac{e ( - 1 ) ^{n}}{π ^{2} n ^{2} + 1}

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\frac{d}{d x} (ln tan (\frac{x}{2}))

\frac{d}{d x} (\frac{9 - x ^{2}}{32 x})

s (t) = 14 t^{2} + 840 t

What is the integral from-1 to 1 of e^{-x}cos(npix) ?
The integral from-1 to 1 of e^{-x}cos(npix) is -((-1)^n)/(e(pi^2n^2+1))+(e(-1)^n)/(pi^2n^2+1)