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Question Number 128610 by john_santu last updated on 08/Jan/21

$${\int }_{-\pi /4}^{\pi /4}\frac{\sec \mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}$$

Commented by liberty last updated on 09/Jan/21

$$\mathrm{I}=-{\int }_{-\pi /4}^{\pi /4}\frac{\sec (-\mathrm{x})}{{\mathrm{e}}^{-\mathrm{x}}+1}\mathrm{d}(-\mathrm{x})$$$$\mathrm{I}={\int }_{-\pi /4}^{\pi /4}\frac{\sec \mathrm{x}}{{\mathrm{e}}^{-\mathrm{x}}+1}\mathrm{dx}=\int {}_{-\pi /4}^{\pi /4}\left(\frac{\sec \mathrm{x}}{{\mathrm{e}}^{-\mathrm{x}}(1+{\mathrm{e}}^{\mathrm{x}})}\right)\mathrm{dx}$$$$\mathrm{I}={\int }_{-\pi /4}^{\pi /4}\frac{{\mathrm{e}}^{\mathrm{x}}\sec \mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}$$$$\mathrm{adding}\mathrm{together}\mathrm{two}\mathrm{equation}$$$$2\mathrm{I}={\int }_{-\pi /4}^{\pi /4}\frac{\sec \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\sec \mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}={\int }_{-\pi /4}^{\pi /4}\sec \mathrm{x}\mathrm{dx}$$$$2\mathrm{I}=2{\int }_0^{\pi /4}\sec \mathrm{x}\mathrm{dx}$$$${\mathrm{I}=\ln \mid \sec \mathrm{x}+\tan \mathrm{x}\mid ]}_0^{\pi /4}=\ln (1+\sqrt{2})$$$$$$

Answered by mathmax by abdo last updated on 09/Jan/21

$$\mathrm{I}={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{\mathrm{cosx}(1+{\mathrm{e}}^{\mathrm{x}})}\mathrm{dx}{=}_{\mathrm{x}=-\mathrm{t}}{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{\mathrm{cost}(1+{\mathrm{e}}^{-\mathrm{t}})}\mathrm{dt}\Rightarrow$$$$2\mathrm{I}={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{\mathrm{cosx}}(\frac{1}{1+{\mathrm{e}}^{\mathrm{x}}}+\frac{1}{1+{\mathrm{e}}^{-\mathrm{x}}})\mathrm{dx}={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{\mathrm{cosx}}\left(\frac{1+{\mathrm{e}}^{-\mathrm{x}}+1+{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}+1}\right)\mathrm{dx}$$$$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\mathrm{dx}}{\mathrm{cosx}}=2{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{dx}}{\mathrm{cosx}}\Rightarrow \mathrm{I}={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{dx}}{\mathrm{cosx}}{=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}$$$${\int }_0^{\sqrt{2}-1}\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2).\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}={\int }_0^{\sqrt{2}-1}(\frac{1}{1-\mathrm{t}}+\frac{1}{1+\mathrm{t}})\mathrm{dt}={[\ln \mid \frac{1+\mathrm{t}}{1-\mathrm{t}}\mid ]}_0^{\sqrt{2}-1}$$$$=\ln \mid \frac{\sqrt{2}}{2-\sqrt{2}}\mid$$

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