∫_(−π/4) ^(π/4) e^(−x) sin x dx =


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Question Number 81664 by zainal tanjung last updated on 14/Feb/20

$\underset{- π / 4}{\int^{π / 4}} e^{- x} \sin x d x =$
Commented by zainal tanjung last updated on 14/Feb/20


Commented by abdomathmax last updated on 14/Feb/20
$I=Im(∫_(−(π/4)) ^(π/4) e^(−x) e^(ix) dx)=Im(∫_(−(π/4)) ^(π/4) e^((−1+i)x) dx) and ∫_(−(π/4)) ^(π/4) e^((−1+i)x) dx=[(1/(−1+i))e^((−1+i)x) ]_(−(π/4)) ^(π/4) =((−1)/(1−i)){ e^((−1+i)(π/4)) −e^(−(−1+i)(π/4)) } =((−1−i)/2){ e^(−(π/4)) ((1/(√2))+(i/(√2)))−e^(π/4) ((1/(√2))−(i/(√2)))} =−((1+i)/2){(1/(√2))(e^(−(π/4)) −e^(π/4) )+(i/(√2))(e^(π/4) +e^(−(π/4)) )} =−(1/(√2))(1+i){ e^(−(π/4)) −e^(π/4) +(e^(π/4) +e^(−(π/4)) )i} =−(1/(√2)){e^(−(π/4)) −e^(π/4) +i(e^(π/4) +e^(−(π/4)) )+i(e^(−(π/4)) −e^(π/4) ) −(e^(π/4) −e^(−(π/4)) )} =−(1/(√2)){2e^(−(π/4)) −2e^(π/4) +i(2e^(−(π/4)) )} ⇒ I =2 e^(−(π/4))$
$I = I m (\int_{- \frac{π}{4}}^{\frac{π}{4}} e^{- x} e^{i x} d x) = I m (\int_{- \frac{π}{4}}^{\frac{π}{4}} e^{(- 1 + i) x} d x)$ $a n d \int_{- \frac{π}{4}}^{\frac{π}{4}} e^{(- 1 + i) x} d x = {[\frac{1}{- 1 + i} e^{(- 1 + i) x}]}_{- \frac{π}{4}}^{\frac{π}{4}}$ $= \frac{- 1}{1 - i} {e^{(- 1 + i) \frac{π}{4}} - e^{- (- 1 + i) \frac{π}{4}}}$ $= \frac{- 1 - i}{2} {e^{- \frac{π}{4}} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) - e^{\frac{π}{4}} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})}$ $= - \frac{1 + i}{2} {\frac{1}{\sqrt{2}} (e^{- \frac{π}{4}} - e^{\frac{π}{4}}) + \frac{i}{\sqrt{2}} (e^{\frac{π}{4}} + e^{- \frac{π}{4}})}$ $= - \frac{1}{\sqrt{2}} (1 + i) {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + (e^{\frac{π}{4}} + e^{- \frac{π}{4}}) i}$ $= - \frac{1}{\sqrt{2}} {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + i (e^{\frac{π}{4}} + e^{- \frac{π}{4}}) + i (e^{- \frac{π}{4}} - e^{\frac{π}{4}})$ $- (e^{\frac{π}{4}} - e^{- \frac{π}{4}})} = - \frac{1}{\sqrt{2}} {2 e^{- \frac{π}{4}} - 2 e^{\frac{π}{4}}$ $+ i (2 e^{- \frac{π}{4}})} \Rightarrow I = 2 e^{- \frac{π}{4}}$
Commented by Tony Lin last updated on 14/Feb/20

$\int e^{- x} s i n x d x$ $= - e^{- x} s i n x + \int e^{- x} c o s x d x$ $= - e^{- x} s i n x + (- e^{- x} c o s x - \int e^{- x} s i n x d x)$ $\Rightarrow \int e^{- x} s i n x d x$ $= \frac{- e^{- x} (s i n x + c o s x)}{2} + c$ $\Rightarrow \int_{- \frac{π}{4}}^{\frac{π}{4}} e^{- x} s i n x d x$ $= {[\frac{- e^{- x} (s i n x + c o s x)}{2}]}_{- \frac{π}{4}}^{\frac{π}{4}}$ $= - \frac{e^{- \frac{π}{4}}}{\sqrt{2}}$
Commented by mathmax by abdo last updated on 14/Feb/20
$sorry ∫_(−(π/4)) ^(π/4) e^((−1+i)x) dx =−(1/(2(√2)))(1+i){e^(−(π/4)) −e^(π/4) +i(e^(π/4) +e^(−(π/4)) )} =−(1/(2(√2))){e^(−(π/4)) −e^(π/4) +i(e^(π/4) +e^(−(π/4)) )+i(e^(−(π/4)) −e^(π/4) )−(e^(π/4) +e^(−(π/4)) )} ⇒ I =−(1/(2(√2)))(2e^(−(π/4)) )=−(1/(√2)) e^(−(π/4)) ★ I=−(1/(√2))e^(−(π/4)) ★$
$s o r r y \int_{- \frac{π}{4}}^{\frac{π}{4}} e^{(- 1 + i) x} d x = - \frac{1}{2 \sqrt{2}} (1 + i) {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + i (e^{\frac{π}{4}} + e^{- \frac{π}{4}})}$ $= - \frac{1}{2 \sqrt{2}} {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + i (e^{\frac{π}{4}} + e^{- \frac{π}{4}}) + i (e^{- \frac{π}{4}} - e^{\frac{π}{4}}) - (e^{\frac{π}{4}} + e^{- \frac{π}{4}})} \Rightarrow$ $I = - \frac{1}{2 \sqrt{2}} (2 e^{- \frac{π}{4}}) = - \frac{1}{\sqrt{2}} e^{- \frac{π}{4}}$ $★ I = - \frac{1}{\sqrt{2}} e^{- \frac{π}{4}} ★$
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