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Question Number 76232 by Rio Michael last updated on 25/Dec/19

$$howdowefind$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}{sinh}^{-1}xdxand\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\cosh {}^{-1}xdx$$

Commented by mr W last updated on 25/Dec/19

$${\cosh }^{-1}xisdefinedforx⩾1!$$

Commented by turbo msup by abdo last updated on 25/Dec/19

$$letI={\int }_0^{\frac{\pi }{2}}argsh\left(x\right)dx$$$$\Rightarrow I{=}_{argsh\left(x\right)=t}{\int }_0^{argsh\left(\frac{\pi }{2}\right)}tch\left(t\right)dt$$$$={\int }_0^{ln(\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}})}tcht\left(\right)dt$$$${=}_{byparts}{\left[tsh\left(t\right)\right]}_0^{ln(\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}})}$$$$-{\int }_0^{ln(\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}})}sh\left(t\right)dt$$$$=ln(\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}})\frac{\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}}-{(\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}})}^{-1}}{2}$$$$-\frac{1}{2}(\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}})+{(\frac{\pi }{2}+\sqrt{1+\frac{{\pi }^2}{4}})}^{-1})$$$$$$

Answered by mind is power last updated on 25/Dec/19

$$\mathrm{u}={\mathrm{sh}}^{-}\left(\mathrm{x}\right)$$$$\mathrm{by}\mathrm{part}$$$$\int {\mathrm{sh}}^{-}\left(\mathrm{x}\right)\mathrm{dx}=\left[{\mathrm{xsh}}^{-}\left(\mathrm{x}\right)\right]-\int \frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}$$$$={\mathrm{xsh}}^{-}\left(\mathrm{x}\right)-\sqrt{{\mathrm{x}}^2+1}+\mathrm{c}$$$$\mathrm{sam}\mathrm{with}{\mathrm{ch}}^{-}\left(\mathrm{x}\right)$$

Answered by mr W last updated on 25/Dec/19

$${\int }_0^{\frac{\pi }{2}}{\sinh }^{-1}xdx$$$$=\frac{\pi }{2}{\sinh }^{-1}\left(\frac{\pi }{2}\right)-{\int }_0^{{\sinh }^{-1}\left(\frac{\pi }{2}\right)}\sinh ydy$$$$=\frac{\pi }{2}{\sinh }^{-1}\left(\frac{\pi }{2}\right)-[\cosh \left({\sinh }^{-1}\frac{\pi }{2}\right)-1]$$$$=\frac{\pi }{2}\ln (\frac{\pi }{2}+\sqrt{1+{\left(\frac{\pi }{2}\right)}^2})+1-\sqrt{1+{\left(\frac{\pi }{2}\right)}^2}$$$$\approx 1.075329$$

Commented by mr W last updated on 25/Dec/19

Commented by mr W last updated on 25/Dec/19

$$I_1={\int }_0^{\frac{\pi }{2}}{\sinh }^{-1}xdx$$$$I_2={\int }_0^{{\sinh }^{-1}\frac{\pi }{2}}\sinh ydy$$$$I_1+I_2=\frac{\pi }{2}\times {\sinh }^{-1}\frac{\pi }{2}$$$$I_{2}iseasiertogetthan{I}_1.$$

Commented by Rio Michael last updated on 25/Dec/19

$$thanks$$

Commented by mr W last updated on 25/Dec/19

$$\text{You can't use 'macro parameter character \#' in math mode}$$

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