Integration Questions

Question Number 61533 by maxmathsup by imad last updated on 04/Jun/19

$$\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\:\:\frac{{x}−{y}}{\left({x}^{\mathrm{2}} \:+\mathrm{3}{y}^{\mathrm{2}\:} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dxdy}\: \\$$

Commented by maxmathsup by imad last updated on 04/Jun/19

$${let}\:{use}\:{the}\:{diffeomorphism}\:\:{x}={rcos}\theta\:{and}\:{y}\:=\frac{{r}}{\sqrt{\mathrm{3}}}\:{sin}\theta\:\: \\$$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}\:\Rightarrow\mathrm{0}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{2}\:\Rightarrow\:\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}} \\$$$$\left({r},\theta\right)\rightarrow\varphi\left({r},\theta\right)\:=\left({x},{y}\right)=\left(\varphi_{\mathrm{1}} \left({r},\theta\right),\varphi_{\mathrm{2}} \left({r},\theta\right)\right)=\left({rcos}\theta,\frac{{r}}{\sqrt{\mathrm{3}}}{sin}\theta\right) \\$$$${M}_{{j}} \left(\varphi\right)\:=\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial\theta}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial\theta}}\end{pmatrix}\:\:\:\:\:\:=\begin{pmatrix}{{cos}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:−{rsin}\theta}\\{\frac{{sin}\theta}{\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{\sqrt{\mathrm{3}}}\:{cos}\theta}\end{pmatrix} \\$$$$\Rightarrow{detM}_{{j}} =\:\frac{{r}}{\sqrt{\mathrm{3}}}\:\Rightarrow{A}=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\:\:\frac{{x}−{y}}{\left({x}^{\mathrm{2}} \:+\mathrm{3}{y}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dxdy}\:=\int\int_{\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{2}}\:{and}\:\:\mathrm{0}\leqslant\theta\:\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{rcos}\theta\:−\frac{{r}}{\sqrt{\mathrm{3}}}{sin}\theta}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\frac{{r}}{\sqrt{\mathrm{3}}}{dr}\:{d}\theta \\$$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{r}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\left({cos}\theta\:−{sin}\theta\right){d}\theta \\$$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{r}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dr}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{r}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dr}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{dr}}{\mathrm{1}+{r}^{\mathrm{2}} }\:−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$={arctan}\left(\sqrt{\mathrm{2}}\right)−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=_{{r}={tanu}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} \:\:\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {u}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right)^{\mathrm{2}} }{du}\:=\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} \:\:\frac{{du}}{\mathrm{1}+{tan}^{\mathrm{2}} {u}} \\$$$$=\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} {cos}^{\mathrm{2}} {u}\:{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} \left(\mathrm{1}+{cos}\left(\mathrm{2}{u}\right)\right){du} \\$$$$=\frac{{arctan}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{u}\right)\right]_{\mathrm{0}} ^{{arctan}\left(\sqrt{\mathrm{2}}\right)} =\frac{{arctan}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\mathrm{2}}\right)\right. \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}\theta\:−{sin}\theta\right){d}\theta\:=\left[{sin}\theta\:+{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{1}−\mathrm{1}\:=\mathrm{0}\:\Rightarrow\:{A}\:=\mathrm{0}\:. \\$$

Commented by maxmathsup by imad last updated on 05/Jun/19

$${error}\:{from}\:{line}\:\mathrm{6}\: \\$$$${I}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\sqrt{\mathrm{3}}{cos}\theta\:−{sin}\theta\right){d}\theta \\$$$$\int_{\mathrm{0}} ^{\frac{\pi{u}}{\mathrm{2}}} \left(\:\sqrt{\mathrm{3}}{cos}\theta\:−{sin}\theta\right){d}\theta\:=\left[\sqrt{\mathrm{3}}{sin}\theta\:+{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\sqrt{\mathrm{3}}−\mathrm{1}\:\Rightarrow \\$$$${I}\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{r}^{\mathrm{2}} {dr}}{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}}\:\left\{\:{arctan}\left(\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\mathrm{2}}\right)\right)\right\} \\$$$$\Rightarrow\:{I}\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}}\left\{\:\frac{{arctan}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\mathrm{2}}\right)\right\}\:.\right. \\$$