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Question Number 126274 by bramlexs22 last updated on 19/Dec/20

Answered by liberty last updated on 19/Dec/20

$$lettingx=\sqrt{2}\sec \ell with\to \{\begin{array}{l}\ell =\frac{\pi }{2}\\ \ell =\frac{\pi }{4}\end{array}$$$${\int }_{\pi /4}^{\pi /2}\frac{\sqrt{2}\sec \ell \tan \ell }{\sqrt{2}\sec \ell \sqrt{2\tan {}^2\ell }}d\ell =$$$$\frac{1}{\sqrt{2}}\int d\ell =\frac{1}{\sqrt{2}}[\frac{\pi }{2}-\frac{\pi }{4}]=\frac{\pi }{4\sqrt{2}}=\frac{\pi \sqrt{2}}{8}$$

Answered by mathmax by abdo last updated on 19/Dec/20

$$\mathrm{I}={\int }_1^{\mathrm{\infty }}\frac{\mathrm{dx}}{\mathrm{x}\sqrt{{\mathrm{x}}^2-2}}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\frac{\sqrt{2}}{\mathrm{sint}}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\frac{\sqrt{2}\mathrm{cost}}{{\sin }^2\mathrm{t}}\Rightarrow$$$${\mathrm{x}}^2-2>0\Rightarrow \mathrm{x}>\sqrt{2}\mathrm{or}\mathrm{x}<-\sqrt{2}\Rightarrow \mathrm{I}\mathrm{is}\mathrm{not}\mathrm{defined}...!$$

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