Question Number 77918 by john santu last updated on 12/Jan/20
$$\int\underset{\mathrm{0}} {\overset{\pi} {\:}}\:{e}^{−\mathrm{2}{x}} \:\mathrm{sin}\:{x}\:{dx}\:?\: \\ $$
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)](https://www.tinkutara.com/question/Q77993.png)
$$\int_{\mathrm{0}} ^{\pi} \:{e}^{−\mathrm{2}{x}} {sinx}\:{dx}\:={Im}\left(\int_{\mathrm{0}} ^{\pi} \:{e}^{−\mathrm{2}{x}+{ix}} {dx}\right)\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{e}^{\left(−\mathrm{2}+{i}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{−\mathrm{2}+{i}}{e}^{\left(−\mathrm{2}+{i}\right){x}} \right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{1}}{−\mathrm{2}+{i}}\left({e}^{\left(−\mathrm{2}+{i}\right)\pi} −\mathrm{1}\right) \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}−{i}}\left\{\:−{e}^{−\mathrm{2}\pi} −\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}−{i}}\left({e}^{−\mathrm{2}\pi} \:+\mathrm{1}\right)\:=\frac{\mathrm{2}+{i}}{\mathrm{5}}\left({e}^{−\mathrm{2}\pi} +\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{1}+{e}^{−\mathrm{2}\pi} \right)+\frac{\mathrm{1}}{\mathrm{5}}\left({e}^{−\mathrm{2}\pi} \:+\mathrm{1}\right){i}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{e}^{−\mathrm{2}{x}} {sinxdx}\:=\frac{\mathrm{1}+{e}^{−\mathrm{2}\pi} }{\mathrm{5}} \\ $$
Answered by john santu last updated on 12/Jan/20
