∫ _0 ^π e^(−2x) sin x dx ?


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Question Number 77918 by john santu last updated on 12/Jan/20

$\int \underset{0}{\overset{π}{}} e^{- 2 x} \sin x d x ?$
Commented by mathmax by abdo last updated on 12/Jan/20
$∫_0 ^π e^(−2x) sinx dx =Im(∫_0 ^π e^(−2x+ix) dx) we have ∫_0 ^π e^((−2+i)x) dx =[(1/(−2+i))e^((−2+i)x) ]_0 ^π =(1/(−2+i))(e^((−2+i)π) −1) =((−1)/(2−i)){ −e^(−2π) −1) =(1/(2−i))(e^(−2π) +1) =((2+i)/5)(e^(−2π) +1) =(2/5)(1+e^(−2π) )+(1/5)(e^(−2π) +1)i ⇒ ∫_0 ^π e^(−2x) sinxdx =((1+e^(−2π) )/5)$
$\int_{0}^{π} e^{- 2 x} s i n x d x = I m (\int_{0}^{π} e^{- 2 x + i x} d x) w e h a v e$ $\int_{0}^{π} e^{(- 2 + i) x} d x = {[\frac{1}{- 2 + i} e^{(- 2 + i) x}]}_{0}^{π} = \frac{1}{- 2 + i} (e^{(- 2 + i) π} - 1)$ $= \frac{- 1}{2 - i} {- e^{- 2 π} - 1) = \frac{1}{2 - i} (e^{- 2 π} + 1) = \frac{2 + i}{5} (e^{- 2 π} + 1)$ $= \frac{2}{5} (1 + e^{- 2 π}) + \frac{1}{5} (e^{- 2 π} + 1) i \Rightarrow$ $\int_{0}^{π} e^{- 2 x} s i n x d x = \frac{1 + e^{- 2 π}}{5}$
	Answered by john santu last updated on 12/Jan/20

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