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+(π/4))+e^(−x) cos(x+(π/4))= 5+(√2)((((e^x +e^(−x) )/2))cosx+(((e^x −e^(−x) )/2))sinx)= 5+(√2)(coshx.cosx+sinhx.sinx)=u ⇒du=2(√2)sinhx.cosx ⇒Ω=∫_(5+(√2)) ^( 5+e^(π/4) ) (((1/(2(√2)))du)/u)=((√2)/4)[ln∣u∣]_(5+(√2)) ^(5+e^(π/4) ) = ((√2)/4)ln(((5+e^(π/4) )/(5+(√2))))](https://www.tinkutara.com/question/Q163394.png)
$$\mathrm{5}+{e}^{{x}} {sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)+{e}^{−{x}} {cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)= \\ $$$$\mathrm{5}+\sqrt{\mathrm{2}}\left(\left(\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}\right){cosx}+\left(\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}\right){sinx}\right)= \\ $$$$\mathrm{5}+\sqrt{\mathrm{2}}\left({coshx}.{cosx}+{sinhx}.{sinx}\right)={u} \\ $$$$\Rightarrow{du}=\mathrm{2}\sqrt{\mathrm{2}}{sinhx}.{cosx} \\ $$$$\Rightarrow\Omega=\int_{\mathrm{5}+\sqrt{\mathrm{2}}} ^{\:\mathrm{5}+{e}^{\pi/\mathrm{4}} } \:\:\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{du}}{{u}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left[{ln}\mid{u}\mid\right]_{\mathrm{5}+\sqrt{\mathrm{2}}} ^{\mathrm{5}+{e}^{\pi/\mathrm{4}} } = \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{ln}\left(\frac{\mathrm{5}+{e}^{\pi/\mathrm{4}} }{\mathrm{5}+\sqrt{\mathrm{2}}}\right) \\ $$
Commented by HongKing last updated on 06/Jan/22
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$