What is the integral from-1 to 1 of e^xcos(npix) ?

The integral from-1 to 1 of e^xcos(npix) is (e(-1)^n)/(pi^2n^2+1)-((-1)^n)/(e(pi^2n^2+1))

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integral from-1 to 1 of e^xcos(npix)

\int_{- 1}^{1} e^{x} cos (nπ x) d x

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

\frac{d}{d t} (a sin (3 t) + b cos (3 t))

n = 1 \sum \infty \frac{1}{5 ^{\frac{1}{n}}}

x \to - 3 lim (\frac{x + 3}{x ^{2} - 9})

f (x) = 2 x e^{x}

\frac{d}{d x} (\frac{x e ^{x}}{x + 1})

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