integral from 0 to pi of (sin(x))/(e^x)

解

\int_{0}^{π} \frac{sin ( x )}{e ^{x}} d x

\frac{1}{2 e ^{π}} + \frac{1}{2}

十進法表記

0.52160 \dots

解答ステップ

\int_{0}^{π} \frac{sin ( x )}{e ^{x}} d x

部分積分を適用する

= [\frac{- sin ( x ) - e ^{x} \cdot \int - \frac{c o s ( x )}{e ^{x}} d x}{e ^{x}}]_{0}^{π}

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

= [\frac{- sin ( x ) - e ^{x} ( - \frac{1}{2} e ^{- x} sin ( x ) + \frac{1}{2} e ^{- x} cos ( x ) )}{e ^{x}}]_{0}^{π}

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

= [\frac{- sin ( x ) - cos ( x )}{2 e ^{x}}]_{0}^{π}

限度を計算する:

\frac{1}{2 e ^{π}} + \frac{1}{2}

= \frac{1}{2 e ^{π}} + \frac{1}{2}