What is the integral from 0 to pi of e^{-inx} ?

The integral from 0 to pi of e^{-inx} is i((-1)^n-1)/n

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integral from 0 to pi of e^{-inx}

\int_{0}^{π} e^{- in x} d x

i \frac{( - 1 ) ^{n} - 1}{n}

Solution steps

\int_{0}^{π} e^{- in x} d x

Simplify

e^{- in x} : cos (n x) - i sin (n x)

= \int_{0}^{π} cos (n x) - i sin (n x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} cos (n x) d x - \int_{0}^{π} i sin (n x) d x

\int_{0}^{π} cos (n x) d x = 0

\int_{0}^{π} i sin (n x) d x = i \frac{- ( - 1 ) ^{n} + 1}{n}

= 0 - i \frac{- ( - 1 ) ^{n} + 1}{n}

Simplify

0 - i \frac{- ( - 1 ) ^{n} + 1}{n} : i \frac{( - 1 ) ^{n} - 1}{n}

= i \frac{( - 1 ) ^{n} - 1}{n}

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\int_{0}^{2 π} \frac{cos ( 2 x )}{2} d x

\int_{0}^{2} (y + 2) d y

What is the integral from 0 to pi of e^{-inx} ?
The integral from 0 to pi of e^{-inx} is i((-1)^n-1)/n