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Question Number 130738 by bemath last updated on 28/Jan/21

$${\int }_1^{\cosh \mathrm{x}}\sqrt{{\mathrm{t}}^2-1}\mathrm{dt}=?$$

Answered by MJS_new last updated on 28/Jan/21

$$\int \sqrt{t^2-1}dt=$$$$[t=\cosh u\iff u={\cosh }^{-1}t\to dt=\sinh udu]$$$$=\int {\sinh }^{2}udu=-\frac{1}{2}(u-\sinh u\cosh u)=$$$$=-\frac{1}{2}({\cosh }^{-1}t-t\sqrt{t^2-1})+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}-\frac{x-\sinh x\cosh x}{2}$$

Commented by bemath last updated on 29/Jan/21

$$\mathrm{great}$$

Answered by mathmax by abdo last updated on 28/Jan/21

$$\mathrm{f}\left(\mathrm{x}\right)={\int }_1^{\mathrm{ch}\left(\mathrm{x}\right)}\sqrt{{\mathrm{t}}^2-1}\mathrm{dt}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{t}=\mathrm{chz}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\mathrm{shz}\mathrm{shz}\mathrm{dz}={\int }_0^{\mathrm{x}}\frac{\mathrm{ch}\left(2\mathrm{z}\right)-1}{2}\mathrm{dz}$$$${=\frac{1}{4}\mathrm{sh}\left(2\mathrm{z}\right)]}_0^{\mathrm{x}}-\frac{\mathrm{x}}{2}=\frac{1}{4}\mathrm{sh}\left(2\mathrm{x}\right)-\frac{\mathrm{x}}{2}.$$

Commented by bemath last updated on 29/Jan/21

$$\mathrm{nice}$$

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