What is the integral from 0 to pi of e^{-px}sin(x) ?

The integral from 0 to pi of e^{-px}sin(x) is (e^{-pip}+1)/(p^2+1)

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integral from 0 to pi of e^{-px}sin(x)

\int_{0}^{π} e^{- p x} sin (x) d x

\frac{e ^{- π p} + 1}{p ^{2} + 1}

Solution steps

\int_{0}^{π} e^{- p x} sin (x) d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

= \frac{e ^{- π p} + 1}{p ^{2} + 1}

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What is the integral from 0 to pi of e^{-px}sin(x) ?
The integral from 0 to pi of e^{-px}sin(x) is (e^{-pip}+1)/(p^2+1)