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Question Number 145044 by mathmax by abdo last updated on 01/Jul/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{x}}\sqrt{1+\mathrm{sinx}}\mathrm{dx}$$

Answered by qaz last updated on 02/Jul/21

$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{x}}\sqrt{1+\sin \mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{x}}(\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2})\mathrm{dx}$$$$=\mathcal{L}(\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2})(\mathrm{s}=2)$$$$=\frac{\frac{1}{2}+s}{{\mathrm{s}}^2+\frac{1}{4}}{\mid }_{\mathrm{s}=2}$$$$=\frac{10}{17}$$

Commented by mathmax by abdo last updated on 03/Jul/21

$$\mathrm{answer}\mathrm{not}\mathrm{correct}\mathrm{sir}\mathrm{gaz}$$

Answered by mnjuly1970 last updated on 02/Jul/21

$$\sqrt{1+sin\left(x\right)}=\mid sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)\mid$$

Answered by mathmax by abdo last updated on 02/Jul/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{x}}\sqrt{1+\mathrm{sinx}}\mathrm{dx}\mathrm{we}\mathrm{have}1+\mathrm{sinx}={(\cos \left(\frac{\mathrm{x}}{2}\right)+\sin \left(\frac{\mathrm{x}}{2}\right))}^2\Rightarrow$$$$\sqrt{1+\mathrm{sinx}}=\mid \cos \left(\frac{\mathrm{x}}{2}\right)+\sin \left(\frac{\mathrm{x}}{2}\right)\mid \Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{x}}\mid \cos \left(\frac{\mathrm{x}}{2}\right)+\sin \left(\frac{\mathrm{x}}{2}\right)\mid \mathrm{dx}{=}_{\frac{\mathrm{x}}{2}=\mathrm{t}}2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{t}}\mid \mathrm{cost}+\mathrm{sint}\mid \mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{t}}\mid \sqrt{2}\sin (\mathrm{t}+\frac{\pi }{4})\mid \mathrm{dt}=2\sqrt{2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{t}}\mid \sin (\mathrm{t}+\frac{\pi }{4})\mid \mathrm{dt}$$$${=}_{\mathrm{t}+\frac{\pi }{4}=\mathrm{y}}2\sqrt{2}{\int }_{\frac{\pi }{4}}^{+\mathrm{\infty }}{\mathrm{e}}^{-4(\mathrm{y}-\frac{\pi }{4})}\mid \mathrm{siny}\mid \mathrm{dy}$$$$=2\sqrt{2}{\mathrm{e}}^{\pi }{\int }_{\frac{\pi }{4}}^{+\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{y}}\mid \mathrm{siny}\mid \mathrm{dy}$$$$=2\sqrt{2}{\mathrm{e}}^{\pi }({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{y}}\mid \mathrm{siny}\mid \mathrm{dy}-{\int }_0^{\frac{\pi }{4}}{\mathrm{e}}^{-4\mathrm{y}}\mathrm{siny}\mathrm{dy})\mathrm{we}\mathrm{have}$$$${\int }_0^{\frac{\pi }{4}}{\mathrm{e}}^{-4\mathrm{y}}\mathrm{siny}\mathrm{dy}=\mathrm{Im}\left({\int }_0^{\frac{\pi }{4}}{\mathrm{e}}^{-4\mathrm{y}+\mathrm{iy}}\mathrm{dy}\right)$$$${\int }_0^{\frac{\pi }{4}}{\mathrm{e}}^{(-4+\mathrm{i})\mathrm{y}}\mathrm{dy}={\left[\frac{1}{-4+\mathrm{i}}{\mathrm{e}}^{(-4+\mathrm{i})\mathrm{y}}\right]}_0^{\frac{\pi }{4}}$$$$=-\frac{1}{4-\mathrm{i}}({\mathrm{e}}^{(-4+\mathrm{i})\frac{\pi }{4}}-1)$$$$=-\frac{4+\mathrm{i}}{17}({\mathrm{e}}^{-\pi }(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})-1)$$$$=-\frac{1}{17}(4+\mathrm{i})(\frac{{\mathrm{e}}^{-\pi }}{\sqrt{2}}+\mathrm{i}\frac{{\mathrm{e}}^{-\pi }}{\sqrt{2}}-1)$$$$=-\frac{1}{17}(\frac{{4\mathrm{e}}^{-\pi }}{\sqrt{2}}+4\mathrm{i}\frac{{\mathrm{e}}^{-\pi }}{\sqrt{2}}-4+\mathrm{i}\frac{{\mathrm{e}}^{-\pi }}{\sqrt{2}}-\frac{{\mathrm{e}}^{-\pi }}{\sqrt{2}}-\mathrm{i})\Rightarrow$$$${\int }_0^{\frac{\pi }{4}}{\mathrm{e}}^{-4\mathrm{y}}\mathrm{sinydy}=-\frac{1}{17}(\frac{{4\mathrm{e}}^{-\pi }}{\sqrt{2}}+\frac{{\mathrm{e}}^{-\pi }}{\sqrt{2}}-1)=\frac{1}{17}(1-\frac{{5\mathrm{e}}^{-\pi }}{\sqrt{2}})$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{y}}\mid \mathrm{siny}\mid \mathrm{dy}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_{\mathrm{n}\pi }^{(\mathrm{n}+1)\pi }{\mathrm{e}}^{-4\mathrm{y}}\mid \mathrm{siny}\mid \mathrm{dy}$$$${=}_{\mathrm{y}=\mathrm{n}\pi +\mathrm{z}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\pi }{\mathrm{e}}^{-4(\mathrm{n}\pi +\mathrm{z})}\mathrm{sinz}\mathrm{dz}(\mathrm{sinz}⩾0\mathrm{on}[0,\pi ])$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{n}\pi }{\int }_0^{\pi }{\mathrm{e}}^{-4\mathrm{z}}\mathrm{sinzdz}\mathrm{we}\mathrm{have}$$$${\int }_0^{\pi }{\mathrm{e}}^{-4\mathrm{z}}\mathrm{sinz}\mathrm{dz}=\mathrm{Im}\left({\int }_0^{\pi }{\mathrm{e}}^{-4\mathrm{z}+\mathrm{iz}}\mathrm{dz}\right)$$$${\int }_0^{\pi }{\mathrm{e}}^{(-4+\mathrm{i})\mathrm{z}}\mathrm{dz}={\left[\frac{1}{-4+\mathrm{i}}{\mathrm{e}}^{(-4+\mathrm{i})\mathrm{z}}\right]}_0^{\pi }$$$$=-\frac{1}{4-\mathrm{i}}({\mathrm{e}}^{(-4+\mathrm{i})\pi }-1)$$$$=-\frac{4+\mathrm{i}}{17}(-{\mathrm{e}}^{-4\pi }-1)=\frac{4+\mathrm{i}}{17}(1+{\mathrm{e}}^{-4\pi })\Rightarrow$$$${\int }_0^{\pi }{\mathrm{e}}^{-4\mathrm{z}}\mathrm{sinz}\mathrm{dz}=\frac{1+{\mathrm{e}}^{-4\pi }}{17}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{y}}\mid \mathrm{siny}\mid \mathrm{dy}=\frac{1+{\mathrm{e}}^{-4\pi }}{17}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left({\mathrm{e}}^{-4\pi }\right)}^{\mathrm{n}}$$$$=\frac{1+{\mathrm{e}}^{-4\pi }}{17}\times \frac{1}{1-{\mathrm{e}}^{-4\pi }}=\frac{1+{\mathrm{e}}^{4\pi }}{17(1-{\mathrm{e}}^{-4\pi })}\Rightarrow$$$$\mathrm{I}=2\sqrt{2}{\mathrm{e}}^{\pi }(\frac{1+{\mathrm{e}}^{4\pi }}{17(1-{\mathrm{e}}^{4\pi })}-\frac{1}{17}(1-\frac{{5\mathrm{e}}^{-\pi }}{\sqrt{2}})$$$$=\frac{2\sqrt{2}}{17}(\frac{1+{\mathrm{e}}^{4\pi }}{1-{\mathrm{e}}^{4\pi }}-1+\frac{{5\mathrm{e}}^{-\pi }}{\sqrt{2}})$$

Commented by mathmax by abdo last updated on 02/Jul/21

$$\mathrm{I}=\frac{2\sqrt{2}}{17}(\frac{1+{\mathrm{e}}^{4\pi }}{1-{\mathrm{e}}^{-4\pi }}-1+\frac{{5\mathrm{e}}^{-\pi }}{\sqrt{2}})$$

Commented by mnjuly1970 last updated on 03/Jul/21

$$verynicesirmax...$$

Commented by qaz last updated on 04/Jul/21

$$\mathrm{nice}\mathrm{solution}\mathrm{sir}$$

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