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 121712 by rs4089 last updated on 11/Nov/20

Answered by Ar Brandon last updated on 11/Nov/20

$${\int }_0^{\frac{\mathrm{a}}{\sqrt{2}}}{\int }_{\mathrm{y}}^{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}\ln ({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{dxdy}$$$$\mathrm{Let}\mathcal{I}={\int }_{\mathrm{y}}^{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}\ln ({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{dx}$$$$\Rightarrow \mathcal{I}={\{\mathrm{xln}({\mathrm{x}}^2+{\mathrm{y}}^2)-2\int \frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{dx}\}}_{\mathrm{y}}^{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}$$$$={\{\mathrm{xln}({\mathrm{x}}^2+{\mathrm{y}}^2)-2\int [1-\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}]\mathrm{dx}\}}_{\mathrm{y}}^{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}$$$$=\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}\ln \left({\mathrm{a}}^2\right)-\mathrm{yln}\left({2\mathrm{y}}^2\right)-2{[\mathrm{x}-\mathrm{Arctan}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)]}_{\mathrm{y}}^{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}$$$$=\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}\ln \left({\mathrm{a}}^2\right)-\mathrm{yln}({2\mathrm{y}}^2-2[\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}-\mathrm{yArctan}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{\mathrm{y}}\right)-\mathrm{y}+\mathrm{yArctan}\left(1\right)]$$$$=({\mathrm{lna}}^2-2)\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}-\mathrm{yln}\left({2\mathrm{y}}^2\right)+2\mathrm{yArctan}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{\mathrm{y}}\right)+2\mathrm{y}-\frac{\pi }{2}\mathrm{y}$$$${\int }_0^{\frac{\mathrm{a}}{\sqrt{2}}}\left(\mathcal{I}\right)\mathrm{dy}={\{({\mathrm{lna}}^2-2)\int \sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}\mathrm{dy}-\int \mathrm{yln}\left({2\mathrm{y}}^2\right)\mathrm{dy}+2\int \mathrm{yArctan}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{\mathrm{y}}\right)\mathrm{dy}+\int (2-\frac{\pi }{2})\mathrm{ydy}\}}_0^{\frac{\mathrm{a}}{\sqrt{2}}}$$

Commented by Ar Brandon last updated on 11/Nov/20

$$\int \sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}\mathrm{dy}$$$$\mathrm{y}=\mathrm{asin}\theta \Rightarrow \mathrm{dy}=\mathrm{acos}\theta \mathrm{d}\theta$$$$=\int \sqrt{{\mathrm{a}}^2-{\mathrm{a}}^2{\sin }^2\theta }\cdot \mathrm{acos}\theta \mathrm{d}\theta$$$$={\mathrm{a}}^2\int {\cos }^2\theta \mathrm{d}\theta =\frac{{\mathrm{a}}^2}{2}\int (1+\cos 2\theta )\mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^2}{2}(\theta +\frac{\sin 2\theta }{2}).\theta =\mathrm{Arcsin}\left(\frac{\mathrm{y}}{\mathrm{a}}\right)$$

Commented by Ar Brandon last updated on 11/Nov/20

$$\int \mathrm{yln}\left({2\mathrm{y}}^2\right)\mathrm{dy}=\frac{1}{4}\int 4\mathrm{yln}\left({2\mathrm{y}}^2\right)\mathrm{dy}$$$$=\frac{{2\mathrm{y}}^2[\ln \left({2\mathrm{y}}^2\right)-1]}{4}$$

Commented by Ar Brandon last updated on 11/Nov/20

$$\int \mathrm{yArctan}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{\mathrm{y}}\right)\mathrm{dy}$$$$\mathrm{u}\left(\mathrm{y}\right)=\mathrm{Arctan}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{\mathrm{y}}\right)$$$${\mathrm{u}}^{\prime }\left(\mathrm{y}\right)=\frac{\mathrm{y}\times (-2\mathrm{y})\times \frac{1}{2}\cdot \frac{1}{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}-\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{{\mathrm{y}}^2}$$$$=-\frac{1}{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}-\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{{\mathrm{y}}^2}$$$${\mathrm{v}}^{\prime }\left(\mathrm{y}\right)=\mathrm{y}\Rightarrow \mathrm{v}\left(\mathrm{y}\right)=\frac{{\mathrm{y}}^2}{2}$$$$\int \mathrm{yArctan}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{\mathrm{y}}\right)\mathrm{dy}$$$$=\frac{{\mathrm{y}}^2}{2}\mathrm{Arctan}\left(\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{\mathrm{y}}\right)+\frac{1}{2}\int [\frac{{\mathrm{y}}^2}{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}+\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}]$$$$\int \frac{{\mathrm{y}}^2}{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}\mathrm{dy}=\int \{\frac{{\mathrm{a}}^2}{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}-\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}\}\mathrm{dy}$$$$\int \frac{{\mathrm{a}}^2}{\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}\mathrm{dy}={\mathrm{a}}^2\mathrm{Arcsin}\left(\frac{\mathrm{y}}{\mathrm{a}}\right)$$

Commented by Ar Brandon last updated on 11/Nov/20

$$\int (2-\frac{\pi }{2})\mathrm{ydy}=(1-\frac{\pi }{4}){\mathrm{y}}^2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com