Evaluating $int_{1/4}^1 int_{sqrt{x-x^2}}^{sqrt x}left(frac{x^2-y^2}{x^2}right),dy,dx$












2












$begingroup$


I am looking for an efficient way to evaluate




$$int_{1/4}^1 int_{sqrt{x-x^2}}^{sqrt x}left(frac{x^2-y^2}{x^2}right),mathrm{d}y,mathrm{d}x$$




I have



begin{align}
I&=int_{1/4}^1left(int_{sqrt{x-x^2}}^{sqrt x},dy-frac{1}{x^2}int_{sqrt{x-x^2}}^{sqrt x}y^2,dyright),dx
\&=int_{1/4}^1left[sqrt x-sqrt{x-x^2}-frac{(sqrt{x})^3-(sqrt{x-x^2})^3}{3x^2}right],dx
\&=int_{1/4}^1sqrt x,dx-int_{1/4}^1 sqrt{x-x^2},dx-frac{1}{3}int_{1/4}^1 frac{1}{sqrt x},dx+frac{1}{3}int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx
\&=frac{2}{3}left(1-frac{1}{4^{3/2}}right)-frac{1}{3}-color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}+frac{1}{3}color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}
end{align}



Using this answer,



$$color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}=frac{1}{4}int_{-pi/6}^{pi/2}cos^2,dt=frac{1}{8}int_{-pi/6}^{pi/2}(1+cos 2t),dt=frac{1}{96}(3sqrt 3+8pi)$$



For $color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}$ or even the indefinite integral, nothing comes to mind off the top of my head.



In a different approach, if I try to change the order of integration right at the start, it complicates matters for me to rewrite the region $$sqrt{x-x^2}<y<sqrt x,,,1/4<x<1$$



keeping a separate range of $y$ free of $x$ and bounding $x$ with $y$.



Any suggestion regarding specific substitution or change of variables would also be helpful.






integration multivariable-calculus multiple-integral







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  • $begingroup$
    symbolab.com/solver/step-by-step/… this link will show you how to integrate your 'green' integral
    $endgroup$
    – ricky
    Dec 19 '18 at 9:37
















2












$begingroup$


I am looking for an efficient way to evaluate




$$int_{1/4}^1 int_{sqrt{x-x^2}}^{sqrt x}left(frac{x^2-y^2}{x^2}right),mathrm{d}y,mathrm{d}x$$




I have



begin{align}
I&=int_{1/4}^1left(int_{sqrt{x-x^2}}^{sqrt x},dy-frac{1}{x^2}int_{sqrt{x-x^2}}^{sqrt x}y^2,dyright),dx
\&=int_{1/4}^1left[sqrt x-sqrt{x-x^2}-frac{(sqrt{x})^3-(sqrt{x-x^2})^3}{3x^2}right],dx
\&=int_{1/4}^1sqrt x,dx-int_{1/4}^1 sqrt{x-x^2},dx-frac{1}{3}int_{1/4}^1 frac{1}{sqrt x},dx+frac{1}{3}int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx
\&=frac{2}{3}left(1-frac{1}{4^{3/2}}right)-frac{1}{3}-color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}+frac{1}{3}color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}
end{align}



Using this answer,



$$color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}=frac{1}{4}int_{-pi/6}^{pi/2}cos^2,dt=frac{1}{8}int_{-pi/6}^{pi/2}(1+cos 2t),dt=frac{1}{96}(3sqrt 3+8pi)$$



For $color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}$ or even the indefinite integral, nothing comes to mind off the top of my head.



In a different approach, if I try to change the order of integration right at the start, it complicates matters for me to rewrite the region $$sqrt{x-x^2}<y<sqrt x,,,1/4<x<1$$



keeping a separate range of $y$ free of $x$ and bounding $x$ with $y$.



Any suggestion regarding specific substitution or change of variables would also be helpful.






integration multivariable-calculus multiple-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    symbolab.com/solver/step-by-step/… this link will show you how to integrate your 'green' integral
    $endgroup$
    – ricky
    Dec 19 '18 at 9:37














2












2








2





$begingroup$


I am looking for an efficient way to evaluate




$$int_{1/4}^1 int_{sqrt{x-x^2}}^{sqrt x}left(frac{x^2-y^2}{x^2}right),mathrm{d}y,mathrm{d}x$$




I have



begin{align}
I&=int_{1/4}^1left(int_{sqrt{x-x^2}}^{sqrt x},dy-frac{1}{x^2}int_{sqrt{x-x^2}}^{sqrt x}y^2,dyright),dx
\&=int_{1/4}^1left[sqrt x-sqrt{x-x^2}-frac{(sqrt{x})^3-(sqrt{x-x^2})^3}{3x^2}right],dx
\&=int_{1/4}^1sqrt x,dx-int_{1/4}^1 sqrt{x-x^2},dx-frac{1}{3}int_{1/4}^1 frac{1}{sqrt x},dx+frac{1}{3}int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx
\&=frac{2}{3}left(1-frac{1}{4^{3/2}}right)-frac{1}{3}-color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}+frac{1}{3}color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}
end{align}



Using this answer,



$$color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}=frac{1}{4}int_{-pi/6}^{pi/2}cos^2,dt=frac{1}{8}int_{-pi/6}^{pi/2}(1+cos 2t),dt=frac{1}{96}(3sqrt 3+8pi)$$



For $color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}$ or even the indefinite integral, nothing comes to mind off the top of my head.



In a different approach, if I try to change the order of integration right at the start, it complicates matters for me to rewrite the region $$sqrt{x-x^2}<y<sqrt x,,,1/4<x<1$$



keeping a separate range of $y$ free of $x$ and bounding $x$ with $y$.



Any suggestion regarding specific substitution or change of variables would also be helpful.






integration multivariable-calculus multiple-integral







share|cite|improve this question











$endgroup$




I am looking for an efficient way to evaluate




$$int_{1/4}^1 int_{sqrt{x-x^2}}^{sqrt x}left(frac{x^2-y^2}{x^2}right),mathrm{d}y,mathrm{d}x$$




I have



begin{align}
I&=int_{1/4}^1left(int_{sqrt{x-x^2}}^{sqrt x},dy-frac{1}{x^2}int_{sqrt{x-x^2}}^{sqrt x}y^2,dyright),dx
\&=int_{1/4}^1left[sqrt x-sqrt{x-x^2}-frac{(sqrt{x})^3-(sqrt{x-x^2})^3}{3x^2}right],dx
\&=int_{1/4}^1sqrt x,dx-int_{1/4}^1 sqrt{x-x^2},dx-frac{1}{3}int_{1/4}^1 frac{1}{sqrt x},dx+frac{1}{3}int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx
\&=frac{2}{3}left(1-frac{1}{4^{3/2}}right)-frac{1}{3}-color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}+frac{1}{3}color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}
end{align}



Using this answer,



$$color{darkred}{int_{1/4}^1 sqrt{x-x^2},dx}=frac{1}{4}int_{-pi/6}^{pi/2}cos^2,dt=frac{1}{8}int_{-pi/6}^{pi/2}(1+cos 2t),dt=frac{1}{96}(3sqrt 3+8pi)$$



For $color{green}{int_{1/4}^1frac{(x-x^2)^{3/2}}{x^2},dx}$ or even the indefinite integral, nothing comes to mind off the top of my head.



In a different approach, if I try to change the order of integration right at the start, it complicates matters for me to rewrite the region $$sqrt{x-x^2}<y<sqrt x,,,1/4<x<1$$



keeping a separate range of $y$ free of $x$ and bounding $x$ with $y$.



Any suggestion regarding specific substitution or change of variables would also be helpful.






integration multivariable-calculus multiple-integral




integration multivariable-calculus multiple-integral






share|cite|improve this question















share|cite|improve this question













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edited Dec 19 '18 at 9:33







StubbornAtom

















asked Dec 19 '18 at 9:28









StubbornAtomStubbornAtom

6,06811239




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  • $begingroup$
    symbolab.com/solver/step-by-step/… this link will show you how to integrate your 'green' integral
    $endgroup$
    – ricky
    Dec 19 '18 at 9:37


















  • $begingroup$
    symbolab.com/solver/step-by-step/… this link will show you how to integrate your 'green' integral
    $endgroup$
    – ricky
    Dec 19 '18 at 9:37
















$begingroup$
symbolab.com/solver/step-by-step/… this link will show you how to integrate your 'green' integral
$endgroup$
– ricky
Dec 19 '18 at 9:37




$begingroup$
symbolab.com/solver/step-by-step/… this link will show you how to integrate your 'green' integral
$endgroup$
– ricky
Dec 19 '18 at 9:37










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The substitution $u=sqrt{x}$ gives



$$ int_{1/4}^1 frac{(x-x^2)^{3/2}}{x^2}dx = int_{1/4}^1 frac{(1-x)^{3/2}}{sqrt{x}}dx = 2int_{1/2}^1 (1-u^2)^{3/2} du $$



Then $u=sin t$ should do the trick.






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    $begingroup$

    The substitution $u=sqrt{x}$ gives



    $$ int_{1/4}^1 frac{(x-x^2)^{3/2}}{x^2}dx = int_{1/4}^1 frac{(1-x)^{3/2}}{sqrt{x}}dx = 2int_{1/2}^1 (1-u^2)^{3/2} du $$



    Then $u=sin t$ should do the trick.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The substitution $u=sqrt{x}$ gives



      $$ int_{1/4}^1 frac{(x-x^2)^{3/2}}{x^2}dx = int_{1/4}^1 frac{(1-x)^{3/2}}{sqrt{x}}dx = 2int_{1/2}^1 (1-u^2)^{3/2} du $$



      Then $u=sin t$ should do the trick.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The substitution $u=sqrt{x}$ gives



        $$ int_{1/4}^1 frac{(x-x^2)^{3/2}}{x^2}dx = int_{1/4}^1 frac{(1-x)^{3/2}}{sqrt{x}}dx = 2int_{1/2}^1 (1-u^2)^{3/2} du $$



        Then $u=sin t$ should do the trick.






        share|cite|improve this answer









        $endgroup$



        The substitution $u=sqrt{x}$ gives



        $$ int_{1/4}^1 frac{(x-x^2)^{3/2}}{x^2}dx = int_{1/4}^1 frac{(1-x)^{3/2}}{sqrt{x}}dx = 2int_{1/2}^1 (1-u^2)^{3/2} du $$



        Then $u=sin t$ should do the trick.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 11:39









        DylanDylan

        13.5k31027




        13.5k31027






























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