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

∫_0 ^(2π) e^(x/2) sin ((x/2)+(π/4))dx = ?

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

Commented by prof Abdo imad last updated on 06/Jul/18

I =Im( ∫_0 ^(2π) e^(x/2) e^(i((x/2)+(π/4))) dx) =Im( ∫_0 ^(2π) e^((1+i)(x/2) +((iπ)/4)) dx) but ∫_0 ^(2π) e^((1+i)(x/2)+((iπ)/4)) dx=e^((iπ)/4) ∫_0 ^(2π) e^(((1+i)/2)x) dx =e^((iπ)/4) [(2/(1+i)) e^(((1+i)/2)x) ]_0 ^(2π) = (2/(1+i))e^((iπ)/4) { e^((1+i)π) −1} =(2/(1+i)) e^((iπ)/4) { −e^π −1} =((2(1−i))/2) {−e^π −1} e^((iπ)/(4 )) =(−1+i)((1/(√2)) +(i/(√2)))(1+e^π ) =(−(1/(√2)) −(i/(√2)) +(i/(√2)) −(1/(√2)))(1+e^π ) =−(√2) (1+e^π ) but I =Im( ∫....) ⇒I=0

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

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

=(1/((√2) ))∫_0 ^(2Π) e^(x/2) ×(sin(x/2)+cos(x/2)) =(2/(√2))∫_0 ^(2Π) {sin(x/2).((d(e^(x/2) ))/dx)+e^(x/2) .((d(sin(x/2)))/dx)}dx =(√2) ∫_0 ^(2Π) (d/dx)(e^(x/2) sin(x/2))dx =(√2) ∣e^(x/2) sin(x/2)∣_0 ^(2Π) =(√2) {e^Π sinΠ−e^0 sin0}=0 pls check

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

Commented by rahul 19 last updated on 06/Jul/18

Thank you sir! ����

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