Evaluate-0-2pi-e-x-2-sin-x-2-pi-4-dx-

Question Number 13622 by Tinkutara last updated on 21/May/17

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

Answered by ajfour last updated on 22/May/17

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

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

l e t t = \frac{x}{2} \Rightarrow d t = \frac{d x}{2}

w h e n x = 0, t = 0;

a n d f o r x = 2 π, t = π

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

= {[e^{t} \sin (\frac{π}{4} + t)]}_{0}^{π} - \int_{0}^{π} e^{t} \cos (\frac{π}{4} + t) d t

= - \frac{1}{\sqrt{2}} (e^{π} + 1) - {[e^{t} \cos (\frac{π}{4} + t)]}_{0}^{π}

+ \int_{0}^{π} e^{t} \sin (\frac{π}{4} + t) d t

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

\Rightarrow I = 0 .

Commented by ajfour last updated on 22/May/17

c a n n o t y o u c o n f i r m t h e a n s w e r . .

Commented by Tinkutara last updated on 22/May/17

Answer given is 0 .

Commented by ajfour last updated on 22/May/17

t h a n k s, n o t i c e t h e d i f f e r e n c e!

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