Integration Questions

Question Number 39431 by rahul 19 last updated on 06/Jul/18

$$\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{e}^{\frac{{x}}{\mathrm{2}}} \mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mathrm{d}{x}\:=\:? \\$$

Commented byprof Abdo imad last updated on 06/Jul/18

$${I}\:={Im}\left(\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{e}^{\frac{{x}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)} {dx}\right) \\$$ $$={Im}\left(\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:{e}^{\left(\mathrm{1}+{i}\right)\frac{{x}}{\mathrm{2}}\:+\frac{{i}\pi}{\mathrm{4}}} {dx}\right)\:{but} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:{e}^{\left(\mathrm{1}+{i}\right)\frac{{x}}{\mathrm{2}}+\frac{{i}\pi}{\mathrm{4}}} {dx}={e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:{e}^{\frac{\mathrm{1}+{i}}{\mathrm{2}}{x}} {dx} \\$$ $$={e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:\left[\frac{\mathrm{2}}{\mathrm{1}+{i}}\:{e}^{\frac{\mathrm{1}+{i}}{\mathrm{2}}{x}} \right]_{\mathrm{0}} ^{\mathrm{2}\pi} =\:\frac{\mathrm{2}}{\mathrm{1}+{i}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\left\{\:{e}^{\left(\mathrm{1}+{i}\right)\pi} \:−\mathrm{1}\right\} \\$$ $$=\frac{\mathrm{2}}{\mathrm{1}+{i}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \left\{\:−{e}^{\pi} \:−\mathrm{1}\right\} \\$$ $$=\frac{\mathrm{2}\left(\mathrm{1}−{i}\right)}{\mathrm{2}}\:\left\{−{e}^{\pi} −\mathrm{1}\right\}\:{e}^{\frac{{i}\pi}{\mathrm{4}\:}} \:=\left(−\mathrm{1}+{i}\right)\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\frac{{i}}{\sqrt{\mathrm{2}}}\right)\left(\mathrm{1}+{e}^{\pi} \right) \\$$ $$=\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:−\frac{{i}}{\sqrt{\mathrm{2}}}\:+\frac{{i}}{\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\left(\mathrm{1}+{e}^{\pi} \right) \\$$ $$=−\sqrt{\mathrm{2}}\:\left(\mathrm{1}+{e}^{\pi} \right)\:\:\:{but}\:{I}\:={Im}\left(\:\int....\right)\:\Rightarrow{I}=\mathrm{0} \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18

$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {e}^{\frac{{x}}{\mathrm{2}}} ×\left({sin}\frac{{x}}{\mathrm{2}}+{cos}\frac{{x}}{\mathrm{2}}\right) \\$$ $$=\frac{\mathrm{2}}{\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \left\{{sin}\frac{{x}}{\mathrm{2}}.\frac{{d}\left({e}^{\frac{{x}}{\mathrm{2}}} \right)}{{dx}}+{e}^{\frac{{x}}{\mathrm{2}}} .\frac{{d}\left({sin}\frac{{x}}{\mathrm{2}}\right)}{{dx}}\right\}{dx} \\$$ $$=\sqrt{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \frac{{d}}{{dx}}\left({e}^{\frac{{x}}{\mathrm{2}}} {sin}\frac{{x}}{\mathrm{2}}\right){dx} \\$$ $$=\sqrt{\mathrm{2}}\:\mid{e}^{\frac{{x}}{\mathrm{2}}} {sin}\frac{{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\mathrm{2}\Pi} =\sqrt{\mathrm{2}}\:\left\{{e}^{\Pi} {sin}\Pi−{e}^{\mathrm{0}} {sin}\mathrm{0}\right\}=\mathrm{0} \\$$ $${pls}\:{check} \\$$

Commented byrahul 19 last updated on 06/Jul/18

Thank you sir! ����