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

\int_{0}^{2 π} e^{\frac{x}{2}} \sin (\frac{x}{2} + \frac{π}{4}) d x = ?

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

I = I m (\int_{0}^{2 π} e^{\frac{x}{2}} e^{i (\frac{x}{2} + \frac{π}{4})} d x)

= I m (\int_{0}^{2 π} e^{(1 + i) \frac{x}{2} + \frac{i π}{4}} d x) b u t

\int_{0}^{2 π} e^{(1 + i) \frac{x}{2} + \frac{i π}{4}} d x = e^{\frac{i π}{4}} \int_{0}^{2 π} e^{\frac{1 + i}{2} x} d x

= e^{\frac{i π}{4}} {[\frac{2}{1 + i} e^{\frac{1 + i}{2} x}]}_{0}^{2 π} = \frac{2}{1 + i} e^{\frac{i π}{4}} {e^{(1 + i) π} - 1}

= \frac{2}{1 + i} e^{\frac{i π}{4}} {- e^{π} - 1}

= \frac{2 (1 - i)}{2} {- e^{π} - 1} e^{\frac{i π}{4}} = (- 1 + i) (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) (1 + e^{π})

= (- \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} + \frac{i}{\sqrt{2}} - \frac{1}{\sqrt{2}}) (1 + e^{π})

= - \sqrt{2} (1 + e^{π}) b u t I = I m (\int \dots .) \Rightarrow I = 0

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

= \frac{1}{\sqrt{2}} \int_{0}^{2 Π} e^{\frac{x}{2}} \times (s i n \frac{x}{2} + c o s \frac{x}{2})

= \frac{2}{\sqrt{2}} \int_{0}^{2 Π} {s i n \frac{x}{2} . \frac{d (e^{\frac{x}{2}})}{d x} + e^{\frac{x}{2}} . \frac{d (s i n \frac{x}{2})}{d x}} d x

= \sqrt{2} \int_{0}^{2 Π} \frac{d}{d x} (e^{\frac{x}{2}} s i n \frac{x}{2}) d x

= \sqrt{2} ∣ e^{\frac{x}{2}} s i n \frac{x}{2} ∣_{0}^{2 Π} = \sqrt{2} {e^{Π} s i n Π - e^{0} s i n 0} = 0

p l s c h e c k

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

Thank you sir! ��