Integration Questions

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 bymathmax by abdo last updated on 12/Jan/20

$$\int_{\mathrm{0}} ^{\pi} \:{e}^{−\mathrm{2}{x}} {sinx}\:{dx}\:={Im}\left(\int_{\mathrm{0}} ^{\pi} \:{e}^{−\mathrm{2}{x}+{ix}} {dx}\right)\:{we}\:{have} \\$$ $$\int_{\mathrm{0}} ^{\pi} \:{e}^{\left(−\mathrm{2}+{i}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{−\mathrm{2}+{i}}{e}^{\left(−\mathrm{2}+{i}\right){x}} \right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{1}}{−\mathrm{2}+{i}}\left({e}^{\left(−\mathrm{2}+{i}\right)\pi} −\mathrm{1}\right) \\$$ $$=\frac{−\mathrm{1}}{\mathrm{2}−{i}}\left\{\:−{e}^{−\mathrm{2}\pi} −\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}−{i}}\left({e}^{−\mathrm{2}\pi} \:+\mathrm{1}\right)\:=\frac{\mathrm{2}+{i}}{\mathrm{5}}\left({e}^{−\mathrm{2}\pi} +\mathrm{1}\right) \\$$ $$=\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{1}+{e}^{−\mathrm{2}\pi} \right)+\frac{\mathrm{1}}{\mathrm{5}}\left({e}^{−\mathrm{2}\pi} \:+\mathrm{1}\right){i}\:\Rightarrow \\$$ $$\int_{\mathrm{0}} ^{\pi} \:{e}^{−\mathrm{2}{x}} {sinxdx}\:=\frac{\mathrm{1}+{e}^{−\mathrm{2}\pi} }{\mathrm{5}} \\$$

Answered by john santu last updated on 12/Jan/20