Question Number 121712 by rs4089 last updated on 11/Nov/20
Answered by Ar Brandon last updated on 11/Nov/20
![∫_0 ^(a/( (√2))) ∫_y ^(√(a^2 −y^2 )) ln(x^2 +y^2 )dxdy Let I=∫_y ^(√(a^2 −y^2 )) ln(x^2 +y^2 )dx ⇒I={xln(x^2 +y^2 )−2∫(x^2 /(x^2 +y^2 ))dx}_y ^(√(a^2 −y^2 )) ={xln(x^2 +y^2 )−2∫[1−(y^2 /(x^2 +y^2 ))]dx}_y ^(√(a^2 −y^2 )) =(√(a^2 −y^2 ))ln(a^2 )−yln(2y^2 )−2[x−Arctan((x/y))]_y ^(√(a^2 −y^2 )) =(√(a^2 −y^2 ))ln(a^2 )−yln(2y^2 −2[(√(a^2 −y^2 ))−yArctan(((√(a^2 −y^2 ))/y))−y+yArctan(1)] =(lna^2 −2)(√(a^2 −y^2 ))−yln(2y^2 )+2yArctan(((√(a^2 −y^2 ))/y))+2y−(π/2)y ∫_0 ^(a/( (√2))) (I)dy={(lna^2 −2)∫(√(a^2 −y^2 ))dy−∫yln(2y^2 )dy+2∫yArctan(((√(a^2 −y^2 ))/y))dy+∫(2−(π/2))ydy}_0 ^(a/( (√2)))](https://www.tinkutara.com/question/Q121755.png)
$$\int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}}} \int_{\mathrm{y}} ^{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$$$\mathrm{Let}\:\mathcal{I}=\int_{\mathrm{y}} ^{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\Rightarrow\mathcal{I}=\left\{\mathrm{xln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)−\mathrm{2}\int\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\mathrm{dx}\right\}_{\mathrm{y}} ^{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:=\left\{\mathrm{xln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)−\mathrm{2}\int\left[\mathrm{1}−\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\right]\mathrm{dx}\right\}_{\mathrm{y}} ^{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{a}^{\mathrm{2}} \right)−\mathrm{yln}\left(\mathrm{2y}^{\mathrm{2}} \right)−\mathrm{2}\left[\mathrm{x}−\mathrm{Arctan}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\right]_{\mathrm{y}} ^{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{a}^{\mathrm{2}} \right)−\mathrm{yln}\left(\mathrm{2y}^{\mathrm{2}} −\mathrm{2}\left[\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }−\mathrm{yArctan}\left(\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}}\right)−\mathrm{y}+\mathrm{yArctan}\left(\mathrm{1}\right)\right]\right. \\ $$$$\:\:\:\:\:\:\:\:=\left(\mathrm{lna}^{\mathrm{2}} −\mathrm{2}\right)\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }−\mathrm{yln}\left(\mathrm{2y}^{\mathrm{2}} \right)+\mathrm{2yArctan}\left(\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}}\right)+\mathrm{2y}−\frac{\pi}{\mathrm{2}}\mathrm{y} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}}} \left(\mathcal{I}\right)\mathrm{dy}=\left\{\left(\mathrm{lna}^{\mathrm{2}} −\mathrm{2}\right)\int\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\mathrm{dy}−\int\mathrm{yln}\left(\mathrm{2y}^{\mathrm{2}} \right)\mathrm{dy}+\mathrm{2}\int\mathrm{yArctan}\left(\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}}\right)\mathrm{dy}+\int\left(\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)\mathrm{ydy}\right\}_{\mathrm{0}} ^{\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}}} \\ $$
Commented by Ar Brandon last updated on 11/Nov/20
$$\int\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\mathrm{dy} \\ $$$$\mathrm{y}=\mathrm{asin}\theta\:\Rightarrow\:\mathrm{dy}=\mathrm{acos}\theta\mathrm{d}\theta \\ $$$$=\int\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta}\centerdot\mathrm{acos}\theta\mathrm{d}\theta \\ $$$$=\mathrm{a}^{\mathrm{2}} \int\mathrm{cos}^{\mathrm{2}} \theta\mathrm{d}\theta=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos2}\theta\right)\mathrm{d}\theta \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\left(\theta+\frac{\mathrm{sin2}\theta}{\mathrm{2}}\right).\:\:\:\theta=\mathrm{Arcsin}\left(\frac{\mathrm{y}}{\mathrm{a}}\right) \\ $$
Commented by Ar Brandon last updated on 11/Nov/20
![∫yln(2y^2 )dy=(1/4)∫4yln(2y^2 )dy =((2y^2 [ln(2y^2 )−1])/4)](https://www.tinkutara.com/question/Q121758.png)
$$\int\mathrm{yln}\left(\mathrm{2y}^{\mathrm{2}} \right)\mathrm{dy}=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{4yln}\left(\mathrm{2y}^{\mathrm{2}} \right)\mathrm{dy} \\ $$$$=\frac{\mathrm{2y}^{\mathrm{2}} \left[\mathrm{ln}\left(\mathrm{2y}^{\mathrm{2}} \right)−\mathrm{1}\right]}{\mathrm{4}} \\ $$
Commented by Ar Brandon last updated on 11/Nov/20
![∫yArctan(((√(a^2 −y^2 ))/y))dy u(y)=Arctan(((√(a^2 −y^2 ))/y)) u′(y)=((y×(−2y)×(1/2)∙(1/( (√(a^2 −y^2 ))))−(√(a^2 −y^2 )))/y^2 ) =−(1/( (√(a^2 −y^2 ))))−((√(a^2 −y^2 ))/y^2 ) v′(y)=y ⇒ v(y)=(y^2 /2) ∫yArctan(((√(a^2 −y^2 ))/y))dy =(y^2 /2)Arctan(((√(a^2 −y^2 ))/y))+(1/2)∫[(y^2 /( (√(a^2 −y^2 ))))+(√(a^2 −y^2 ))] ∫(y^2 /( (√(a^2 −y^2 ))))dy=∫{(a^2 /( (√(a^2 −y^2 ))))−(√(a^2 −y^2 ))}dy ∫(a^2 /( (√(a^2 −y^2 ))))dy=a^2 Arcsin((y/a))](https://www.tinkutara.com/question/Q121760.png)
$$\int\mathrm{yArctan}\left(\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}}\right)\mathrm{dy} \\ $$$$\mathrm{u}\left(\mathrm{y}\right)=\mathrm{Arctan}\left(\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}}\right) \\ $$$$\mathrm{u}'\left(\mathrm{y}\right)=\frac{\mathrm{y}×\left(−\mathrm{2y}\right)×\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}−\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}^{\mathrm{2}} } \\ $$$$\mathrm{v}'\left(\mathrm{y}\right)=\mathrm{y}\:\Rightarrow\:\mathrm{v}\left(\mathrm{y}\right)=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\int\mathrm{yArctan}\left(\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}}\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}\mathrm{Arctan}\left(\frac{\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{y}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int\left[\frac{\mathrm{y}^{\mathrm{2}} }{\:\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}+\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\right] \\ $$$$\int\frac{\mathrm{y}^{\mathrm{2}} }{\:\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}\mathrm{dy}=\int\left\{\frac{\mathrm{a}^{\mathrm{2}} }{\:\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\right\}\mathrm{dy} \\ $$$$\int\frac{\mathrm{a}^{\mathrm{2}} }{\:\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}\mathrm{dy}=\mathrm{a}^{\mathrm{2}} \mathrm{Arcsin}\left(\frac{\mathrm{y}}{\mathrm{a}}\right) \\ $$
Commented by Ar Brandon last updated on 11/Nov/20
$$\int\left(\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)\mathrm{ydy}=\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right)\mathrm{y}^{\mathrm{2}} \\ $$