integral from 0 to pi of e^xsin(nx)

解答

\int_{0}^{π} e^{x} sin (n x) d x

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

求解步骤

\int_{0}^{π} e^{x} sin (n x) d x

计算不定积分:

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

带入上下限计算后得:

\int_{0}^{π} e^{x} sin (n x) d x = - \frac{e ^{π} ( - 1 ) ^{n} n}{n ^{2} + 1} - (- \frac{n}{n ^{2} + 1})

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

化简

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

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