Integral with sine inside log












0












$begingroup$


I want to evaluate the value of the following integral:
$$int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx$$



i have tried 2 methods, but fail to proceed.



1.
begin{align*}
I&=int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx\
frac{dI}{da}&=int_0^frac{pi}{2}frac{2}{1-a^2*sin^4{a}}dx\
&=int_0^frac{pi}{2}frac{1}{1-a*sin^2{x}}+frac{1}{1+a*sin^2{x}}dx
end{align*}



2.
using $u=ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}$ and $dv=frac{dx}{sin^2{x}}$
we have
begin{align*}
I&=bigg[-cot{x}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}bigg]_0^frac{pi}{2}+4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx\
&=4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx
end{align*}
which is similar to the first one






definite-integrals







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    0












    $begingroup$


    I want to evaluate the value of the following integral:
    $$int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx$$



    i have tried 2 methods, but fail to proceed.



    1.
    begin{align*}
    I&=int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx\
    frac{dI}{da}&=int_0^frac{pi}{2}frac{2}{1-a^2*sin^4{a}}dx\
    &=int_0^frac{pi}{2}frac{1}{1-a*sin^2{x}}+frac{1}{1+a*sin^2{x}}dx
    end{align*}



    2.
    using $u=ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}$ and $dv=frac{dx}{sin^2{x}}$
    we have
    begin{align*}
    I&=bigg[-cot{x}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}bigg]_0^frac{pi}{2}+4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx\
    &=4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx
    end{align*}
    which is similar to the first one






    definite-integrals







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      0





      $begingroup$


      I want to evaluate the value of the following integral:
      $$int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx$$



      i have tried 2 methods, but fail to proceed.



      1.
      begin{align*}
      I&=int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx\
      frac{dI}{da}&=int_0^frac{pi}{2}frac{2}{1-a^2*sin^4{a}}dx\
      &=int_0^frac{pi}{2}frac{1}{1-a*sin^2{x}}+frac{1}{1+a*sin^2{x}}dx
      end{align*}



      2.
      using $u=ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}$ and $dv=frac{dx}{sin^2{x}}$
      we have
      begin{align*}
      I&=bigg[-cot{x}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}bigg]_0^frac{pi}{2}+4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx\
      &=4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx
      end{align*}
      which is similar to the first one






      definite-integrals







      share|cite|improve this question









      $endgroup$




      I want to evaluate the value of the following integral:
      $$int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx$$



      i have tried 2 methods, but fail to proceed.



      1.
      begin{align*}
      I&=int_0^frac{pi}{2}frac{1}{sin^2{x}}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}dx\
      frac{dI}{da}&=int_0^frac{pi}{2}frac{2}{1-a^2*sin^4{a}}dx\
      &=int_0^frac{pi}{2}frac{1}{1-a*sin^2{x}}+frac{1}{1+a*sin^2{x}}dx
      end{align*}



      2.
      using $u=ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}$ and $dv=frac{dx}{sin^2{x}}$
      we have
      begin{align*}
      I&=bigg[-cot{x}ln{frac{1+a*sin^2{x}}{1-a*sin^2{x}}}bigg]_0^frac{pi}{2}+4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx\
      &=4aint_0^frac{pi}{2}frac{cos^2{x}}{1-a^2*sin^4{x}}dx
      end{align*}
      which is similar to the first one






      definite-integrals




      definite-integrals






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      asked Jan 24 at 8:09









      Reynan HenryReynan Henry

      751




      751






















          2 Answers
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          $begingroup$

          Try this:



          $$t=cot x;quad dt=-frac{dx}{sin^2 x};frac{1}{sin^2 x}=1+t^2$$



          $$I=int_0^{pi/2}frac{1}{sin^2 x}lnfrac{1+asin^2 x}{1-asin^2 x}dx$$
          $$I=-int_{infty}^{0}lnfrac{1+asin^2 x}{1-asin^2 x}dt=
          int_0^inftylnfrac{1/sin^2 x+a}{1/sin^2 x-a}dt=
          int_0^inftylnfrac{1+t^2+a}{1+t^2-a}dt
          $$



          Now you can consider separately just the indefinite integral $intln(t^2+b)dt$ by integration by parts, or proceed with differentiation with respect to $a$ (and then further proceeding, knowing that $I(a=0)=0$).






          share|cite|improve this answer









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            0












            $begingroup$

            Considering
            $$I=intfrac{dx}{1-a sin ^2(x)}$$ let
            $$tan(x)=t implies x=tan ^{-1}(t) implies dx=frac{dt}{1+t^2}$$ which make
            $$I=intfrac{dt}{1-(a-1) t^2}=frac{tanh ^{-1}left(sqrt{a-1} tright)}{sqrt{a-1}}$$ Just do the same for the other






            share|cite|improve this answer









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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Try this:



              $$t=cot x;quad dt=-frac{dx}{sin^2 x};frac{1}{sin^2 x}=1+t^2$$



              $$I=int_0^{pi/2}frac{1}{sin^2 x}lnfrac{1+asin^2 x}{1-asin^2 x}dx$$
              $$I=-int_{infty}^{0}lnfrac{1+asin^2 x}{1-asin^2 x}dt=
              int_0^inftylnfrac{1/sin^2 x+a}{1/sin^2 x-a}dt=
              int_0^inftylnfrac{1+t^2+a}{1+t^2-a}dt
              $$



              Now you can consider separately just the indefinite integral $intln(t^2+b)dt$ by integration by parts, or proceed with differentiation with respect to $a$ (and then further proceeding, knowing that $I(a=0)=0$).






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Try this:



                $$t=cot x;quad dt=-frac{dx}{sin^2 x};frac{1}{sin^2 x}=1+t^2$$



                $$I=int_0^{pi/2}frac{1}{sin^2 x}lnfrac{1+asin^2 x}{1-asin^2 x}dx$$
                $$I=-int_{infty}^{0}lnfrac{1+asin^2 x}{1-asin^2 x}dt=
                int_0^inftylnfrac{1/sin^2 x+a}{1/sin^2 x-a}dt=
                int_0^inftylnfrac{1+t^2+a}{1+t^2-a}dt
                $$



                Now you can consider separately just the indefinite integral $intln(t^2+b)dt$ by integration by parts, or proceed with differentiation with respect to $a$ (and then further proceeding, knowing that $I(a=0)=0$).






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Try this:



                  $$t=cot x;quad dt=-frac{dx}{sin^2 x};frac{1}{sin^2 x}=1+t^2$$



                  $$I=int_0^{pi/2}frac{1}{sin^2 x}lnfrac{1+asin^2 x}{1-asin^2 x}dx$$
                  $$I=-int_{infty}^{0}lnfrac{1+asin^2 x}{1-asin^2 x}dt=
                  int_0^inftylnfrac{1/sin^2 x+a}{1/sin^2 x-a}dt=
                  int_0^inftylnfrac{1+t^2+a}{1+t^2-a}dt
                  $$



                  Now you can consider separately just the indefinite integral $intln(t^2+b)dt$ by integration by parts, or proceed with differentiation with respect to $a$ (and then further proceeding, knowing that $I(a=0)=0$).






                  share|cite|improve this answer









                  $endgroup$



                  Try this:



                  $$t=cot x;quad dt=-frac{dx}{sin^2 x};frac{1}{sin^2 x}=1+t^2$$



                  $$I=int_0^{pi/2}frac{1}{sin^2 x}lnfrac{1+asin^2 x}{1-asin^2 x}dx$$
                  $$I=-int_{infty}^{0}lnfrac{1+asin^2 x}{1-asin^2 x}dt=
                  int_0^inftylnfrac{1/sin^2 x+a}{1/sin^2 x-a}dt=
                  int_0^inftylnfrac{1+t^2+a}{1+t^2-a}dt
                  $$



                  Now you can consider separately just the indefinite integral $intln(t^2+b)dt$ by integration by parts, or proceed with differentiation with respect to $a$ (and then further proceeding, knowing that $I(a=0)=0$).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 24 at 8:30









                  orionorion

                  13.7k11837




                  13.7k11837























                      0












                      $begingroup$

                      Considering
                      $$I=intfrac{dx}{1-a sin ^2(x)}$$ let
                      $$tan(x)=t implies x=tan ^{-1}(t) implies dx=frac{dt}{1+t^2}$$ which make
                      $$I=intfrac{dt}{1-(a-1) t^2}=frac{tanh ^{-1}left(sqrt{a-1} tright)}{sqrt{a-1}}$$ Just do the same for the other






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Considering
                        $$I=intfrac{dx}{1-a sin ^2(x)}$$ let
                        $$tan(x)=t implies x=tan ^{-1}(t) implies dx=frac{dt}{1+t^2}$$ which make
                        $$I=intfrac{dt}{1-(a-1) t^2}=frac{tanh ^{-1}left(sqrt{a-1} tright)}{sqrt{a-1}}$$ Just do the same for the other






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Considering
                          $$I=intfrac{dx}{1-a sin ^2(x)}$$ let
                          $$tan(x)=t implies x=tan ^{-1}(t) implies dx=frac{dt}{1+t^2}$$ which make
                          $$I=intfrac{dt}{1-(a-1) t^2}=frac{tanh ^{-1}left(sqrt{a-1} tright)}{sqrt{a-1}}$$ Just do the same for the other






                          share|cite|improve this answer









                          $endgroup$



                          Considering
                          $$I=intfrac{dx}{1-a sin ^2(x)}$$ let
                          $$tan(x)=t implies x=tan ^{-1}(t) implies dx=frac{dt}{1+t^2}$$ which make
                          $$I=intfrac{dt}{1-(a-1) t^2}=frac{tanh ^{-1}left(sqrt{a-1} tright)}{sqrt{a-1}}$$ Just do the same for the other







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 24 at 8:30









                          Claude LeiboviciClaude Leibovici

                          124k1157135




                          124k1157135






























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