Solving the integral :$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin2 x}}dx$












2












$begingroup$


Solving the integral :
$$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}dx=?$$




My try:



$$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}cdot frac{1-sqrt{sin 2x}}{1-sqrt{sin 2x}}dx$$



$$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{1-sin 2x}dx$$



$$(sin x-cos x)^2=1-sin 2x$$



So :



$$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{(sin x-cos x)^2}dx$$



Now what?



$$$$






calculus integration definite-integrals







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    2












    $begingroup$


    Solving the integral :
    $$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}dx=?$$




    My try:



    $$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}cdot frac{1-sqrt{sin 2x}}{1-sqrt{sin 2x}}dx$$



    $$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{1-sin 2x}dx$$



    $$(sin x-cos x)^2=1-sin 2x$$



    So :



    $$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{(sin x-cos x)^2}dx$$



    Now what?



    $$$$






    calculus integration definite-integrals







    share|cite|improve this question











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      2












      2








      2


      1



      $begingroup$


      Solving the integral :
      $$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}dx=?$$




      My try:



      $$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}cdot frac{1-sqrt{sin 2x}}{1-sqrt{sin 2x}}dx$$



      $$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{1-sin 2x}dx$$



      $$(sin x-cos x)^2=1-sin 2x$$



      So :



      $$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{(sin x-cos x)^2}dx$$



      Now what?



      $$$$






      calculus integration definite-integrals







      share|cite|improve this question











      $endgroup$




      Solving the integral :
      $$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}dx=?$$




      My try:



      $$int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}}cdot frac{1-sqrt{sin 2x}}{1-sqrt{sin 2x}}dx$$



      $$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{1-sin 2x}dx$$



      $$(sin x-cos x)^2=1-sin 2x$$



      So :



      $$int_0^{frac{pi}{2}}frac{sin x(1-sqrt{sin 2x})}{(sin x-cos x)^2}dx$$



      Now what?



      $$$$






      calculus integration definite-integrals




      calculus integration definite-integrals






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      edited Mar 2 at 9:02









      Zacky

      7,59511061




      7,59511061










      asked Oct 27 '17 at 19:47









      Almot1960Almot1960

      2,312824




      2,312824






















          2 Answers
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          $begingroup$

          The given integral equals
          $$ frac{1}{2}int_{0}^{pi}frac{sintfrac{x}{2}}{1+sqrt{sin x}},dx =frac{1}{2sqrt{2}}int_{0}^{pi}frac{sqrt{1-cos x}}{1+sqrt{sin x}}\=frac{1}{2sqrt{2}}int_{0}^{pi/2}frac{sqrt{1-cos x}+sqrt{1+cos x}}{1+sqrt{sin x}},dx\
          =frac{1}{2}int_{0}^{pi/2}frac{sqrt{1+sin x}}{1+sqrt{sin x}},dx$$
          or
          $$ frac{1}{2}int_{0}^{1}frac{sqrt{1+x}}{sqrt{1-x^2}(1+sqrt{x})},dx = frac{1}{2}int_{0}^{1}frac{dx}{(1+sqrt{x})sqrt{1-x}}=int_{0}^{1}frac{x,dx}{(1+x)sqrt{1-x^2}}$$
          which equals $frac{pi}{2}-int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}$. On the other hand
          $$ int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}=int_{0}^{pi/2}frac{dtheta}{1+sintheta}=int_{0}^{pi/2}frac{dtheta}{1+costheta}=int_{0}^{pi/2}frac{dtheta}{2cos^2tfrac{theta}{2}}=1, $$
          hence:
          $$boxed{int_{0}^{pi/2}frac{sin x}{1+sqrt{sin(2x)}},dx = color{blue}{frac{pi}{2}-1},} $$
          no particular elliptic integral like $Kleft(frac{1}{2}right)$ or $Eleft(frac{1}{2}right)$ is really involved.






          share|cite|improve this answer









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          • $begingroup$
            Nice answer! (y)
            $endgroup$
            – Darío A. Gutiérrez
            Oct 28 '17 at 18:00



















          2












          $begingroup$

          $$I=int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}},dx$$ Using the fact that: $$int_a^b f(x) dx=frac{1}{2} int_a^b (f(x)+f(a+b-x))dx$$ $$Rightarrow I=frac12 int_0^frac{pi}{2} frac{sin x+cos x}{1+sqrt{sin 2x}}dx$$Now let's substitute $,sin x-cos x=u,$ $rightarrow u^2=1-sin(2x)$$rightarrowsin(2x)=1-u^2$ $$I=int_{0}^{1}frac{du}{1+sqrt{1-u^2}}$$ Letting $u=sin t$ yields: $$I=int_0^{frac{pi}{2}}frac{cos t}{1+cos t},dt= int_0^{frac{pi}{2}}left(frac{1+cos t}{1+cos t}-frac{1}{1+cos t}right),dt$$ $$Rightarrow I=frac{pi}{2}-frac{1}{2}int_0^frac{pi}{2}frac{1}{cos^2frac{t}{2}},dt=frac{pi}{2}-1$$






          share|cite|improve this answer











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            2 Answers
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            5












            $begingroup$

            The given integral equals
            $$ frac{1}{2}int_{0}^{pi}frac{sintfrac{x}{2}}{1+sqrt{sin x}},dx =frac{1}{2sqrt{2}}int_{0}^{pi}frac{sqrt{1-cos x}}{1+sqrt{sin x}}\=frac{1}{2sqrt{2}}int_{0}^{pi/2}frac{sqrt{1-cos x}+sqrt{1+cos x}}{1+sqrt{sin x}},dx\
            =frac{1}{2}int_{0}^{pi/2}frac{sqrt{1+sin x}}{1+sqrt{sin x}},dx$$
            or
            $$ frac{1}{2}int_{0}^{1}frac{sqrt{1+x}}{sqrt{1-x^2}(1+sqrt{x})},dx = frac{1}{2}int_{0}^{1}frac{dx}{(1+sqrt{x})sqrt{1-x}}=int_{0}^{1}frac{x,dx}{(1+x)sqrt{1-x^2}}$$
            which equals $frac{pi}{2}-int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}$. On the other hand
            $$ int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}=int_{0}^{pi/2}frac{dtheta}{1+sintheta}=int_{0}^{pi/2}frac{dtheta}{1+costheta}=int_{0}^{pi/2}frac{dtheta}{2cos^2tfrac{theta}{2}}=1, $$
            hence:
            $$boxed{int_{0}^{pi/2}frac{sin x}{1+sqrt{sin(2x)}},dx = color{blue}{frac{pi}{2}-1},} $$
            no particular elliptic integral like $Kleft(frac{1}{2}right)$ or $Eleft(frac{1}{2}right)$ is really involved.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice answer! (y)
              $endgroup$
              – Darío A. Gutiérrez
              Oct 28 '17 at 18:00
















            5












            $begingroup$

            The given integral equals
            $$ frac{1}{2}int_{0}^{pi}frac{sintfrac{x}{2}}{1+sqrt{sin x}},dx =frac{1}{2sqrt{2}}int_{0}^{pi}frac{sqrt{1-cos x}}{1+sqrt{sin x}}\=frac{1}{2sqrt{2}}int_{0}^{pi/2}frac{sqrt{1-cos x}+sqrt{1+cos x}}{1+sqrt{sin x}},dx\
            =frac{1}{2}int_{0}^{pi/2}frac{sqrt{1+sin x}}{1+sqrt{sin x}},dx$$
            or
            $$ frac{1}{2}int_{0}^{1}frac{sqrt{1+x}}{sqrt{1-x^2}(1+sqrt{x})},dx = frac{1}{2}int_{0}^{1}frac{dx}{(1+sqrt{x})sqrt{1-x}}=int_{0}^{1}frac{x,dx}{(1+x)sqrt{1-x^2}}$$
            which equals $frac{pi}{2}-int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}$. On the other hand
            $$ int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}=int_{0}^{pi/2}frac{dtheta}{1+sintheta}=int_{0}^{pi/2}frac{dtheta}{1+costheta}=int_{0}^{pi/2}frac{dtheta}{2cos^2tfrac{theta}{2}}=1, $$
            hence:
            $$boxed{int_{0}^{pi/2}frac{sin x}{1+sqrt{sin(2x)}},dx = color{blue}{frac{pi}{2}-1},} $$
            no particular elliptic integral like $Kleft(frac{1}{2}right)$ or $Eleft(frac{1}{2}right)$ is really involved.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice answer! (y)
              $endgroup$
              – Darío A. Gutiérrez
              Oct 28 '17 at 18:00














            5












            5








            5





            $begingroup$

            The given integral equals
            $$ frac{1}{2}int_{0}^{pi}frac{sintfrac{x}{2}}{1+sqrt{sin x}},dx =frac{1}{2sqrt{2}}int_{0}^{pi}frac{sqrt{1-cos x}}{1+sqrt{sin x}}\=frac{1}{2sqrt{2}}int_{0}^{pi/2}frac{sqrt{1-cos x}+sqrt{1+cos x}}{1+sqrt{sin x}},dx\
            =frac{1}{2}int_{0}^{pi/2}frac{sqrt{1+sin x}}{1+sqrt{sin x}},dx$$
            or
            $$ frac{1}{2}int_{0}^{1}frac{sqrt{1+x}}{sqrt{1-x^2}(1+sqrt{x})},dx = frac{1}{2}int_{0}^{1}frac{dx}{(1+sqrt{x})sqrt{1-x}}=int_{0}^{1}frac{x,dx}{(1+x)sqrt{1-x^2}}$$
            which equals $frac{pi}{2}-int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}$. On the other hand
            $$ int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}=int_{0}^{pi/2}frac{dtheta}{1+sintheta}=int_{0}^{pi/2}frac{dtheta}{1+costheta}=int_{0}^{pi/2}frac{dtheta}{2cos^2tfrac{theta}{2}}=1, $$
            hence:
            $$boxed{int_{0}^{pi/2}frac{sin x}{1+sqrt{sin(2x)}},dx = color{blue}{frac{pi}{2}-1},} $$
            no particular elliptic integral like $Kleft(frac{1}{2}right)$ or $Eleft(frac{1}{2}right)$ is really involved.






            share|cite|improve this answer









            $endgroup$



            The given integral equals
            $$ frac{1}{2}int_{0}^{pi}frac{sintfrac{x}{2}}{1+sqrt{sin x}},dx =frac{1}{2sqrt{2}}int_{0}^{pi}frac{sqrt{1-cos x}}{1+sqrt{sin x}}\=frac{1}{2sqrt{2}}int_{0}^{pi/2}frac{sqrt{1-cos x}+sqrt{1+cos x}}{1+sqrt{sin x}},dx\
            =frac{1}{2}int_{0}^{pi/2}frac{sqrt{1+sin x}}{1+sqrt{sin x}},dx$$
            or
            $$ frac{1}{2}int_{0}^{1}frac{sqrt{1+x}}{sqrt{1-x^2}(1+sqrt{x})},dx = frac{1}{2}int_{0}^{1}frac{dx}{(1+sqrt{x})sqrt{1-x}}=int_{0}^{1}frac{x,dx}{(1+x)sqrt{1-x^2}}$$
            which equals $frac{pi}{2}-int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}$. On the other hand
            $$ int_{0}^{1}frac{dx}{(1+x)sqrt{1-x^2}}=int_{0}^{pi/2}frac{dtheta}{1+sintheta}=int_{0}^{pi/2}frac{dtheta}{1+costheta}=int_{0}^{pi/2}frac{dtheta}{2cos^2tfrac{theta}{2}}=1, $$
            hence:
            $$boxed{int_{0}^{pi/2}frac{sin x}{1+sqrt{sin(2x)}},dx = color{blue}{frac{pi}{2}-1},} $$
            no particular elliptic integral like $Kleft(frac{1}{2}right)$ or $Eleft(frac{1}{2}right)$ is really involved.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 27 '17 at 21:39









            Jack D'AurizioJack D'Aurizio

            291k33284666




            291k33284666












            • $begingroup$
              Nice answer! (y)
              $endgroup$
              – Darío A. Gutiérrez
              Oct 28 '17 at 18:00


















            • $begingroup$
              Nice answer! (y)
              $endgroup$
              – Darío A. Gutiérrez
              Oct 28 '17 at 18:00
















            $begingroup$
            Nice answer! (y)
            $endgroup$
            – Darío A. Gutiérrez
            Oct 28 '17 at 18:00




            $begingroup$
            Nice answer! (y)
            $endgroup$
            – Darío A. Gutiérrez
            Oct 28 '17 at 18:00











            2












            $begingroup$

            $$I=int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}},dx$$ Using the fact that: $$int_a^b f(x) dx=frac{1}{2} int_a^b (f(x)+f(a+b-x))dx$$ $$Rightarrow I=frac12 int_0^frac{pi}{2} frac{sin x+cos x}{1+sqrt{sin 2x}}dx$$Now let's substitute $,sin x-cos x=u,$ $rightarrow u^2=1-sin(2x)$$rightarrowsin(2x)=1-u^2$ $$I=int_{0}^{1}frac{du}{1+sqrt{1-u^2}}$$ Letting $u=sin t$ yields: $$I=int_0^{frac{pi}{2}}frac{cos t}{1+cos t},dt= int_0^{frac{pi}{2}}left(frac{1+cos t}{1+cos t}-frac{1}{1+cos t}right),dt$$ $$Rightarrow I=frac{pi}{2}-frac{1}{2}int_0^frac{pi}{2}frac{1}{cos^2frac{t}{2}},dt=frac{pi}{2}-1$$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              $$I=int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}},dx$$ Using the fact that: $$int_a^b f(x) dx=frac{1}{2} int_a^b (f(x)+f(a+b-x))dx$$ $$Rightarrow I=frac12 int_0^frac{pi}{2} frac{sin x+cos x}{1+sqrt{sin 2x}}dx$$Now let's substitute $,sin x-cos x=u,$ $rightarrow u^2=1-sin(2x)$$rightarrowsin(2x)=1-u^2$ $$I=int_{0}^{1}frac{du}{1+sqrt{1-u^2}}$$ Letting $u=sin t$ yields: $$I=int_0^{frac{pi}{2}}frac{cos t}{1+cos t},dt= int_0^{frac{pi}{2}}left(frac{1+cos t}{1+cos t}-frac{1}{1+cos t}right),dt$$ $$Rightarrow I=frac{pi}{2}-frac{1}{2}int_0^frac{pi}{2}frac{1}{cos^2frac{t}{2}},dt=frac{pi}{2}-1$$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                $$I=int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}},dx$$ Using the fact that: $$int_a^b f(x) dx=frac{1}{2} int_a^b (f(x)+f(a+b-x))dx$$ $$Rightarrow I=frac12 int_0^frac{pi}{2} frac{sin x+cos x}{1+sqrt{sin 2x}}dx$$Now let's substitute $,sin x-cos x=u,$ $rightarrow u^2=1-sin(2x)$$rightarrowsin(2x)=1-u^2$ $$I=int_{0}^{1}frac{du}{1+sqrt{1-u^2}}$$ Letting $u=sin t$ yields: $$I=int_0^{frac{pi}{2}}frac{cos t}{1+cos t},dt= int_0^{frac{pi}{2}}left(frac{1+cos t}{1+cos t}-frac{1}{1+cos t}right),dt$$ $$Rightarrow I=frac{pi}{2}-frac{1}{2}int_0^frac{pi}{2}frac{1}{cos^2frac{t}{2}},dt=frac{pi}{2}-1$$






                share|cite|improve this answer











                $endgroup$



                $$I=int_0^{frac{pi}{2}}frac{sin x}{1+sqrt{sin 2x}},dx$$ Using the fact that: $$int_a^b f(x) dx=frac{1}{2} int_a^b (f(x)+f(a+b-x))dx$$ $$Rightarrow I=frac12 int_0^frac{pi}{2} frac{sin x+cos x}{1+sqrt{sin 2x}}dx$$Now let's substitute $,sin x-cos x=u,$ $rightarrow u^2=1-sin(2x)$$rightarrowsin(2x)=1-u^2$ $$I=int_{0}^{1}frac{du}{1+sqrt{1-u^2}}$$ Letting $u=sin t$ yields: $$I=int_0^{frac{pi}{2}}frac{cos t}{1+cos t},dt= int_0^{frac{pi}{2}}left(frac{1+cos t}{1+cos t}-frac{1}{1+cos t}right),dt$$ $$Rightarrow I=frac{pi}{2}-frac{1}{2}int_0^frac{pi}{2}frac{1}{cos^2frac{t}{2}},dt=frac{pi}{2}-1$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 23 '18 at 18:58

























                answered May 13 '18 at 11:45









                ZackyZacky

                7,59511061




                7,59511061






























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