Question Number 62203 by maxmathsup by imad last updated on 17/Jun/19
![calculate ∫∫_([0,2]^2 ) ((arctan((√(x^2 +y^2 ))))/(3−(√(x^2 +y^2 ))))dxdy](Q62203.png)
$${calculate}\:\int\int_{\left[\mathrm{0},\mathrm{2}\right]^{\mathrm{2}} } \:\:\:\:\frac{{arctan}\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)}{\mathrm{3}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}{dxdy} \\ $$
Commented by maxmathsup by imad last updated on 19/Jun/19
![let A =∫∫_([0,2]^2 ) ((arctan((√(x^2 +y^2 ))))/(3−(√(x^2 +y^2 ))))dxdy let consider the diffeomorphism x =rcosθ and y =rsinθ we have 0≤x≤2 and 0≤y≤2 ⇒0≤x^2 +y^2 ≤8 ⇒ 0≤r^2 ≤8 ⇒0≤r≤2(√2) ⇒ A =∫∫_(0≤r≤2(√2)and 0≤θ≤(π/2)) ((arctan(r))/(3−r)) r drdθ =(π/2) ∫_0 ^(2(√2)) ((rarctan(r))/(3−r))dr ∫_0 ^(2(√2)) ((r arctan(r))/(3−r)) =_(3−r =x) ∫_3 ^(3−2(√2)) (((3−x) arctan(3−x))/x)(−dx) = ∫_(3−2(√2)) ^3 ((3/x)−1)arctan(3−x)dx = 3 ∫_(3−2(√2)) ^3 ((arctan(3−x))/x)dx −∫_(3−2(√2)) ^3 arctan(3−x)dx ∫_(3−2(√2)) ^3 arctan(3−x)dx =_(3−x =t) ∫_(2(√2)) ^0 arctan(t)(−dt) =∫_0 ^(2(√2)) arctan(t)dt =_(by parts) [t arctan(t)]_0 ^(2(√2)) −∫_0 ^(2(√2)) (t/(1+t^2 ))dt=2(√2)arctan(2(√2))−(1/2)[ln(1+t^2 )]_0 ^(2(√2)) =2(√2)arctan(2(√2)) −(1/2)(2ln(3)) = 2(√2)arctan(2(√2))−ln(3).let find ∫_(3−2(√2)) ^3 ((arctan(3−x))/x) dx let f(t) =∫_(3−2(√2)) ^3 ((arctan(3−tx))/x)dx with t≥0 f^′ (t) = ∫_(3−2(√2)) ^3 ((−x)/(x(1+(3−tx)^2 )))dx =−∫_(3−2(√2)) ^3 (dx/(1+9−6tx +t^2 x^2 )) =−∫_(3−2(√2)) ^3 (dx/(t^2 x^2 −6tx +10)) Δ^′ =(−3t)^2 −10t^2 =−t^2 <0 ⇒f^′ (t) =∫_(3−2(√2)) ^3 (dx/(1+(tx−3)^2 )) =_(tx−3 =α) ∫_(t(3−2(√2))−3) ^(3t−3) (dα/(t(1 +α^2 ))) =(1/t)[ arctan(α)]_(t(3−2(√2))−3) ^(3t−3) =(1/t){ arctan(3t−3)−arctan((3−2(√2))t−3)} ⇒ f(t) =∫ ((arctan(3t−3)−arctan{(3−2(√2))t−3})/t) dt +c ...be contnued..](Q62304.png)
$${let}\:\:{A}\:=\int\int_{\left[\mathrm{0},\mathrm{2}\right]^{\mathrm{2}} } \:\:\:\frac{{arctan}\left(\sqrt{\left.{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)}\right.}{\mathrm{3}−\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}{dxdy}\:\:{let}\:{consider}\:{the}\:{diffeomorphism} \\ $$$${x}\:={rcos}\theta\:\:{and}\:{y}\:={rsin}\theta\:\:{we}\:{have}\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\:\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{2}\:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{8}\:\Rightarrow \\ $$$$\mathrm{0}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{8}\:\Rightarrow\mathrm{0}\leqslant{r}\leqslant\mathrm{2}\sqrt{\mathrm{2}}\:\Rightarrow\:{A}\:=\int\int_{\mathrm{0}\leqslant{r}\leqslant\mathrm{2}\sqrt{\mathrm{2}}{and}\:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{arctan}\left({r}\right)}{\mathrm{3}−{r}}\:{r}\:{drd}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\sqrt{\mathrm{2}}} \:\:\frac{{rarctan}\left({r}\right)}{\mathrm{3}−{r}}{dr} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\sqrt{\mathrm{2}}} \:\:\frac{{r}\:{arctan}\left({r}\right)}{\mathrm{3}−{r}}\:=_{\mathrm{3}−{r}\:={x}} \:\:\:\:\int_{\mathrm{3}} ^{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} \:\:\frac{\left(\mathrm{3}−{x}\right)\:{arctan}\left(\mathrm{3}−{x}\right)}{{x}}\left(−{dx}\right) \\ $$$$=\:\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\:\left(\frac{\mathrm{3}}{{x}}−\mathrm{1}\right){arctan}\left(\mathrm{3}−{x}\right){dx}\:=\:\mathrm{3}\:\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\frac{{arctan}\left(\mathrm{3}−{x}\right)}{{x}}{dx} \\ $$$$−\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:{arctan}\left(\mathrm{3}−{x}\right){dx} \\ $$$$\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:{arctan}\left(\mathrm{3}−{x}\right){dx}\:=_{\mathrm{3}−{x}\:={t}} \:\:\:\:\int_{\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{0}} \:{arctan}\left({t}\right)\left(−{dt}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\sqrt{\mathrm{2}}} \:{arctan}\left({t}\right){dt} \\ $$$$=_{{by}\:{parts}} \:\:\:\:\left[{t}\:{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{2}\sqrt{\mathrm{2}}} \:−\int_{\mathrm{0}} ^{\mathrm{2}\sqrt{\mathrm{2}}} \frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{2}\sqrt{\mathrm{2}}{arctan}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}{arctan}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{ln}\left(\mathrm{3}\right)\right)\:=\:\mathrm{2}\sqrt{\mathrm{2}}{arctan}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)−{ln}\left(\mathrm{3}\right).{let}\:{find}\: \\ $$$$\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\frac{{arctan}\left(\mathrm{3}−{x}\right)}{{x}}\:{dx}\:\:{let}\:{f}\left({t}\right)\:=\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\frac{{arctan}\left(\mathrm{3}−{tx}\right)}{{x}}{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$$${f}^{'} \left({t}\right)\:=\:\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\frac{−{x}}{{x}\left(\mathrm{1}+\left(\mathrm{3}−{tx}\right)^{\mathrm{2}} \right)}{dx}\:=−\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\:\frac{{dx}}{\mathrm{1}+\mathrm{9}−\mathrm{6}{tx}\:+{t}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$=−\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\:\:\frac{{dx}}{{t}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{6}{tx}\:+\mathrm{10}} \\ $$$$\Delta^{'} \:=\left(−\mathrm{3}{t}\right)^{\mathrm{2}} −\mathrm{10}{t}^{\mathrm{2}} \:=−{t}^{\mathrm{2}} <\mathrm{0}\:\:\:\:\Rightarrow{f}^{'} \left({t}\right)\:=\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{3}} \:\:\frac{{dx}}{\mathrm{1}+\left({tx}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=_{{tx}−\mathrm{3}\:=\alpha} \:\:\:\:\:\:\:\int_{{t}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\mathrm{3}} ^{\mathrm{3}{t}−\mathrm{3}} \:\:\:\frac{{d}\alpha}{{t}\left(\mathrm{1}\:+\alpha^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{t}}\left[\:{arctan}\left(\alpha\right)\right]_{{t}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\mathrm{3}} ^{\mathrm{3}{t}−\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{{t}}\left\{\:{arctan}\left(\mathrm{3}{t}−\mathrm{3}\right)−{arctan}\left(\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){t}−\mathrm{3}\right)\right\}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\int\:\frac{{arctan}\left(\mathrm{3}{t}−\mathrm{3}\right)−{arctan}\left\{\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){t}−\mathrm{3}\right\}}{{t}}\:{dt}\:+{c}\:\:...{be}\:{contnued}.. \\ $$