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Question Number 163385 by HongKing last updated on 06/Jan/22

Answered by mahdipoor last updated on 06/Jan/22

$$5+e^{x}sin(x+\frac{\pi }{4})+e^{-x}cos(x+\frac{\pi }{4})=$$$$5+\sqrt{2}(\left(\frac{e^x+e^{-x}}{2}\right)cosx+\left(\frac{e^x-e^{-x}}{2}\right)sinx)=$$$$5+\sqrt{2}(coshx.cosx+sinhx.sinx)=u$$$$\Rightarrow du=2\sqrt{2}sinhx.cosx$$$$\Rightarrow \mathrm{\Omega }={\int }_{5+\sqrt{2}}^{5+e^{\pi /4}}\frac{\frac{1}{2\sqrt{2}}du}{u}=\frac{\sqrt{2}}{4}{[ln\mid u\mid ]}_{5+\sqrt{2}}^{5+e^{\pi /4}}=$$$$\frac{\sqrt{2}}{4}ln\left(\frac{5+e^{\pi /4}}{5+\sqrt{2}}\right)$$

Commented by HongKing last updated on 06/Jan/22

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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