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

$$\mathrm{Evaluate}:{\int }_0^{2\pi }{e}^{\frac{x}{2}}\sin (\frac{x}{2}+\frac{\pi }{4})dx$$

Answered by ajfour last updated on 22/May/17

$$I={\int }_0^{2\pi }{e}^{\mathit{x}/2}\sin (\frac{x}{2}+\frac{\pi }{4})dx$$$$\frac{I}{2}={\int }_0^{2\pi }{e}^{\mathit{x}/2}\sin (\frac{x}{2}+\frac{\pi }{4})\frac{dx}{2}$$$$lett=\frac{x}{2}\Rightarrow dt=\frac{dx}{2}$$$$whenx=0,t=0;$$$$andforx=2\pi ,t=\pi$$$$\frac{I}{2}={\int }_0^{\pi }{e}^t\sin (\frac{\pi }{4}+t)dt$$$$={\left[e^t\sin (\frac{\pi }{4}+t)\right]}_0^{\pi }-{\int }_0^{\pi }{e}^t\cos (\frac{\pi }{4}+t)dt$$$$=-\frac{1}{\sqrt{2}}(e^{\pi }+1)-{\left[e^t\cos (\frac{\pi }{4}+t)\right]}_0^{\pi }$$$$+{\int }_0^{\pi }{e}^t\sin (\frac{\pi }{4}+t)dt$$$$\Rightarrow \frac{I}{2}=-\frac{1}{\sqrt{2}}(e^{\pi }+1)+\frac{1}{\sqrt{2}}(e^{\pi }+1)+I$$$$\Rightarrow \mathit{I}=0.$$

Commented by ajfour last updated on 22/May/17

$$cannotyouconfirmtheanswer..$$

Commented by Tinkutara last updated on 22/May/17

$$\mathrm{Answer}\mathrm{given}\mathrm{is}0.$$

Commented by ajfour last updated on 22/May/17

$$thanks,noticethedifference!$$

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