Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 104264 by mohammad17 last updated on 20/Jul/20

$${\int }_0^{\sqrt{2}}{\int }_y^{\sqrt{4-y^2}}\frac{1}{\sqrt{1+x^2+y^2}}dxdy$$

Answered by Ar Brandon last updated on 20/Jul/20

$$\mathcal{I}={\int }_0^{\sqrt{2}}\left\{{\int }_{\mathrm{y}}^{\sqrt{4-{\mathrm{y}}^2}}\frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^2+{\mathrm{y}}^2}}\right\}\mathrm{dy}={\int }_0^{\sqrt{2}}\frac{1}{\sqrt{1+{\mathrm{y}}^2}}\cdot \left\{{\int }_{\mathrm{y}}^{\sqrt{4-{\mathrm{y}}^2}}\frac{\mathrm{dx}}{\sqrt{1+{\left(\frac{\mathrm{x}}{\sqrt{1+{\mathrm{y}}^2}}\right)}^2}}\right\}\mathrm{dy}$$$$={\int }_0^{\sqrt{2}}\frac{1}{\sqrt{1+{\mathrm{y}}^2}}\cdot \frac{\sqrt{1+{\mathrm{y}}^2}}{1}{\left\{\mathrm{Arcsinh}\left(\frac{\mathrm{x}}{\sqrt{1+{\mathrm{y}}^2}}\right)\right\}}_{\mathrm{y}}^{\sqrt{4-{\mathrm{y}}^2}}\mathrm{dy}$$$$={\int }_0^{\sqrt{2}}\{\mathrm{Arcsinh}\left(\sqrt{\frac{4-{\mathrm{y}}^2}{1+{\mathrm{y}}^2}}\right)-\mathrm{Arcsinh}\left(\frac{\mathrm{y}}{\sqrt{1+{\mathrm{y}}^2}}\right)\}\mathrm{dy}$$$$Tobecontinued...$$

Commented by mohammad17 last updated on 20/Jul/20

$$sirbypolar$$

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

$$$$$$\mathrm{En}\mathrm{coordonn}\acute{\mathrm{e}es}\mathrm{polaires}\mathrm{x}=\mathrm{rcos}\theta ,\mathrm{y}=\mathrm{rsin}\theta$$$$0⩽\mathrm{y}⩽\sqrt{2}\Rightarrow 0⩽\mathrm{rsin}\theta ⩽\sqrt{2}\Rightarrow 0⩽\mathrm{r}⩽\frac{\sqrt{2}}{\sin \theta }$$$$\mathrm{y}⩽\mathrm{x}⩽\sqrt{4-{\mathrm{y}}^2}\Rightarrow \{\begin{array}{l}\mathrm{y}⩽\mathrm{x}\\ \mathrm{x}⩽\sqrt{4-{\mathrm{y}}^2}\end{array}\Rightarrow \{\begin{array}{l}\tan \theta ⩽1\\ {\mathrm{x}}^2+{\mathrm{y}}^2⩽4\end{array}\Rightarrow \{\begin{array}{l}\theta ⩽\frac{\pi }{4}\\ \mathrm{r}⩽2\end{array}$$$$\mathrm{r}=2\Rightarrow 0⩽2\sin \theta ⩽\sqrt{2}\Rightarrow 0⩽\theta ⩽\frac{\pi }{4}$$$$\Rightarrow \mathcal{I}={\int }_0^{\frac{\pi }{4}}{\int }_0^{\frac{\sqrt{2}}{\sin \theta }}\frac{\mathrm{r}}{\sqrt{1+{\mathrm{r}}^2}}\mathrm{drd}\theta ={\int }_0^{\frac{\pi }{4}}{\left[\sqrt{1+{\mathrm{r}}^2}\right]}_0^{\frac{\sqrt{2}}{\sin \theta }}\mathrm{d}\theta$$$$\Rightarrow \mathcal{I}={\int }_0^{\frac{\pi }{4}}\{\sqrt{1+\frac{2}{{\sin }^2\theta }}-1\}\mathrm{d}\theta ={\int }_0^{\frac{\pi }{4}}\left\{\sqrt{1+\frac{2}{{\sin }^2\theta }}\right\}\mathrm{d}\theta -\frac{\pi }{4}$$$$$$$$\stackrel{`}{\mathrm{A}}\mathrm{suivre}...$$

Commented by Ar Brandon 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

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

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

$$Withmr{mathmax}^{\prime }shelp$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com