What is the integral from 0 to pi of (sin(x))/(e^x) ?

The integral from 0 to pi of (sin(x))/(e^x) is 1/(2e^pi)+1/2

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

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

\frac{1}{2 e ^{π}} + \frac{1}{2}

Decimal

0.52160 \dots

Solution steps

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

Apply Integration By Parts

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

= [\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}

Compute the boundaries:

\frac{1}{2 e ^{π}} + \frac{1}{2}

= \frac{1}{2 e ^{π}} + \frac{1}{2}

\int_{0}^{\frac{π}{2}} \frac{sin ( 3 x )}{e ^{2 x}} d x

\int_{1}^{2} x (x + 1) d x

\int_{- 7}^{7} (4 + x + x^{2} + x^{3}) d x

\int_{0}^{2} \frac{1}{x - 1} d x

\int_{1}^{3} (\frac{x ^{2} + 5}{x}) d x

What is the integral from 0 to pi of (sin(x))/(e^x) ?
The integral from 0 to pi of (sin(x))/(e^x) is 1/(2e^pi)+1/2