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Question Number 104312 by Ar Brandon last updated on 20/Jul/20

$${\int }_0^{\frac{\pi }{4}}\frac{\sqrt{{\sin }^2\theta +2}}{\sin \theta }\mathrm{d}\theta$$

Answered by mathmax by abdo last updated on 21/Jul/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{4}}\frac{\sqrt{{\sin }^2\theta +2}}{\sin \theta }\mathrm{d}\theta \Rightarrow \mathrm{I}={\int }_0^{\frac{\pi }{4}}\frac{\sin \theta \sqrt{1+\frac{2}{{\sin }^2\theta }}}{\sin \theta }\mathrm{d}\theta ={\int }_0^{\frac{\pi }{4}}\sqrt{1+\frac{2}{1-{\cos }^2\theta }}\mathrm{d}\theta$$$$\mathrm{we}\mathrm{have}1+{\tan }^2\theta =\frac{1}{{\cos }^2\theta }\Rightarrow {\cos }^2\theta =\frac{1}{1+{\tan }^2\theta }\Rightarrow 1-{\cos }^2\theta =1-\frac{1}{1+{\tan }^2\theta }$$$$=\frac{{\tan }^2\theta }{1+{\tan }^2\theta }\Rightarrow \frac{2}{1-{\cos }^2\theta }=2\times \frac{1+{\tan }^2\theta }{{\tan }^2\theta }\Rightarrow$$$$\mathrm{I}={\int }_0^{\frac{\pi }{4}}\sqrt{1+\frac{2+{2\tan }^2\theta }{{\tan }^2\theta }}\mathrm{d}\theta ={\int }_0^{\frac{\pi }{4}}\sqrt{\frac{{3\tan }^2\theta +2}{{\tan }^2\theta }}\mathrm{d}\theta {=}_{\tan \theta =\mathrm{x}}$$$$={\int }_0^1\sqrt{\frac{{3\mathrm{x}}^2+2}{{\mathrm{x}}^2}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}={\int }_0^1\frac{\sqrt{{3\mathrm{x}}^2+2}}{\mathrm{x}(1+{\mathrm{x}}^2)}\mathrm{dx}=\sqrt{3}{\int }_0^1\frac{\sqrt{{\mathrm{x}}^2+\frac{2}{3}}}{\mathrm{x}(1+{\mathrm{x}}^2)}\mathrm{dx}$$$$\mathrm{cha}7\mathrm{gement}\mathrm{x}=\sqrt{\frac{2}{3}}\mathrm{shu}\mathrm{give}\mathrm{u}=\mathrm{argsh}\left(\sqrt{\frac{3}{2}}\mathrm{x}\right)$$$$\mathrm{I}=\sqrt{3}{\int }_0^{\mathrm{arhsh}\left(\sqrt{\frac{3}{2}}\right)}\frac{2}{3}\times \frac{\mathrm{chu}}{\sqrt{\frac{2}{3}}\mathrm{sh}\left(\mathrm{u}\right)(1+\frac{2}{3}{\mathrm{sh}}^2\mathrm{u})}\times \sqrt{\frac{2}{3}}\mathrm{ch}\left(\mathrm{u}\right)\mathrm{du}$$$$=2\sqrt{3}{\int }_0^{\ln (\sqrt{\frac{3}{2}}+\sqrt{1+\frac{9}{4}})}\frac{{\mathrm{ch}}^2\mathrm{u}}{\mathrm{shu}(3+{2\mathrm{sh}}^2\mathrm{u})}\mathrm{du}\mathrm{and}\mathrm{this}\mathrm{integral}\mathrm{can}\mathrm{be}\mathrm{solved}$$$$...\mathrm{be}\mathrm{continued}...$$

Commented by Ar Brandon last updated on 21/Jul/20

wow wow wow ! superb. Thanks so much ��

Commented by Ar Brandon last updated on 21/Jul/20

$$$$$$$$$$\frac{{\mathrm{ch}}^2\mathrm{u}}{\mathrm{shu}(3+{2\mathrm{sh}}^2\mathrm{u})}=\frac{1+{\mathrm{sh}}^2\mathrm{u}}{\mathrm{shu}(3+{2\mathrm{sh}}^2\mathrm{u})}$$$$\frac{1+{\mathrm{t}}^2}{\mathrm{t}(3+{2\mathrm{t}}^2)}=\frac{\mathrm{at}+\mathrm{b}}{{2\mathrm{t}}^2+3}+\frac{\mathrm{c}}{\mathrm{t}}=\frac{(\mathrm{at}+\mathrm{b})\mathrm{t}+\mathrm{c}({2\mathrm{t}}^2+3)}{\mathrm{t}({2\mathrm{t}}^2+3)}$$$$\mathrm{t}\to 0,\mathrm{c}=\frac{1}{3},\mathrm{a}+2\mathrm{c}=1,\mathrm{a}=\frac{1}{3},\mathrm{b}=0$$$$\Rightarrow \frac{1+{\mathrm{sh}}^2\mathrm{u}}{\mathrm{shu}(3+{2\mathrm{sh}}^2\mathrm{u})}=\frac{\frac{1}{3}\mathrm{shu}}{3+{2\mathrm{sh}}^2\mathrm{u}}+\frac{\frac{1}{3}}{\mathrm{shu}}$$$$\Rightarrow \int \frac{{\mathrm{ch}}^2\mathrm{u}}{\mathrm{shu}(3+{2\mathrm{sh}}^2\mathrm{u})}\mathrm{du}=\frac{1}{3}\{\int \frac{\mathrm{shu}\mathrm{du}}{3+{2\mathrm{ch}}^2-2}+\int \frac{\mathrm{du}}{\mathrm{shu}}\}$$$$=\frac{1}{3}\{\frac{1}{\sqrt{2}}\int \frac{\mathrm{d}\left(\sqrt{2}\mathrm{chu}\right)}{1+{2\mathrm{ch}}^2\mathrm{u}}+\int \frac{\mathrm{shu}\mathrm{du}}{{\mathrm{ch}}^2\mathrm{u}-1}\}$$$$=\frac{1}{3}\{\frac{1}{\sqrt{2}}\mathrm{Arctan}\left(\sqrt{2}\mathrm{chu}\right)-\mathrm{Arctanh}\left(\mathrm{chu}\right)\}$$$$$$

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