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Question Number 29105 by tawa tawa last updated on 04/Feb/18

$$\mathrm{Show}\mathrm{that}:{\int }_{-1}^1\frac{\mathrm{dx}}{5\cosh \left(\mathrm{x}\right)+13\sinh \left(\mathrm{x}\right)}=\frac{1}{2}{\log }_{\mathrm{e}}\left(\frac{15\mathrm{e}-10}{3\mathrm{e}+2}\right)$$

Commented by abdo imad last updated on 04/Feb/18

$$I={\int }_{-1}^1\frac{dx}{5\frac{e^x+e^{-x}}{2}+13\frac{e^x-e^{-x}}{2}}$$$$={\int }_{-1}^1\frac{2dx}{18e^x-8e^{-x}}thech.e^x=tgive$$$$I={\int }_{e^{-1}}^e\frac{2}{18t-\frac{8}{t}}\frac{dt}{t}={\int }_{e^{-1}}^e\frac{2}{18t^2-8}dt$$$$I={\int }_{e^{-1}}^e\frac{dt}{9t^2-4}=>9I=\frac{3}{4}{\int }_{e^{-1}}^e(\frac{1}{t-\frac{2}{3}}-\frac{1}{t+\frac{2}{3}})dt$$$$I=\frac{1}{12}ln\mid \frac{t-\frac{2}{3}}{t+\frac{2}{3}}{\mid }_{e^{-1}}^e=\frac{1}{12}(ln\mid \frac{3e-2}{3e+2}\mid -ln\mid \frac{3e^{-1}-2}{3e^{-1}+2}\mid ....$$$$$$

Commented by tawa tawa last updated on 04/Feb/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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