find value of trig integral












4












$begingroup$



Finding
$$int^{pi/2}_0 frac{sin^{n-2}(x)}{(1+cos x)^n}dx$$




what i try



$$
begin{split}
I &= int^{pi/2}_0 left(frac{sin x}{1+cos x}right)^ncsc^2(x)dx \
&= int^{pi/2}_0 tan^n(x/2)csc^2(x)dx
end{split}
$$



But I don't know how to proceed further. Could you please help?






definite-integrals trigonometric-integrals







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$endgroup$

















    4












    $begingroup$



    Finding
    $$int^{pi/2}_0 frac{sin^{n-2}(x)}{(1+cos x)^n}dx$$




    what i try



    $$
    begin{split}
    I &= int^{pi/2}_0 left(frac{sin x}{1+cos x}right)^ncsc^2(x)dx \
    &= int^{pi/2}_0 tan^n(x/2)csc^2(x)dx
    end{split}
    $$



    But I don't know how to proceed further. Could you please help?






    definite-integrals trigonometric-integrals







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$



      Finding
      $$int^{pi/2}_0 frac{sin^{n-2}(x)}{(1+cos x)^n}dx$$




      what i try



      $$
      begin{split}
      I &= int^{pi/2}_0 left(frac{sin x}{1+cos x}right)^ncsc^2(x)dx \
      &= int^{pi/2}_0 tan^n(x/2)csc^2(x)dx
      end{split}
      $$



      But I don't know how to proceed further. Could you please help?






      definite-integrals trigonometric-integrals







      share|cite|improve this question











      $endgroup$





      Finding
      $$int^{pi/2}_0 frac{sin^{n-2}(x)}{(1+cos x)^n}dx$$




      what i try



      $$
      begin{split}
      I &= int^{pi/2}_0 left(frac{sin x}{1+cos x}right)^ncsc^2(x)dx \
      &= int^{pi/2}_0 tan^n(x/2)csc^2(x)dx
      end{split}
      $$



      But I don't know how to proceed further. Could you please help?






      definite-integrals trigonometric-integrals




      definite-integrals trigonometric-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 5 hours ago









      gt6989b

      34k22455




      34k22455










      asked 5 hours ago









      jackyjacky

      709512




      709512






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          Note that the $csc^2(x)$ term can be expressed in terms of $tanleft(frac x2right)$ aswell. To be precise we got that



          begin{align*}
          csc^2(x)=left(frac1{sin(x)}right)^2=left(frac1{2sinleft(frac x2right)cosleft(frac x2right)}right)^2=left(frac{frac1{cos^2left(frac x2right)}}{2frac{sinleft(frac x2right)}{cosleft(frac x2right)}}right)^2
          =left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2
          end{align*}



          Using this and further noticing that $frac{mathrm d}{mathrm dx}tanleft(frac x2right)=frac12left(1+tan^2left(frac x2right)right)$ we may enforce the substition $tanleft(frac x2right)=u$ to obtain



          begin{align*}
          I_n=int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx&=int_0^{pi/2}tan^nleft(frac x2right)left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2mathrm dx\
          &=frac12int_0^{pi/2}tan^{n-2}left(frac x2right)left(1+tan^2left(frac x2right)right)left[frac12left(1+tan^2left(frac x2right)right)mathrm dxright]\
          &=frac12int_0^1 u^{n-2}(1+u^2)mathrm du\
          &=frac12left[frac{u^{n-1}}{n-1}+frac{u^{n+1}}{n+1}right]_0^1\
          &=frac12left[frac1{n-1}+frac1{n+1}right]
          end{align*}




          $$therefore~I_n~=~int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx~=~frac n{n^2-1}$$







          share|cite|improve this answer









          $endgroup$





















            5












            $begingroup$

            Slightly different approach, same answer.



            Let $u=tan(x/2)$, then
            $$
            begin{align}
            int_0^{pi/2}frac{sin^{n-2}(x)}{(1+cos(x))^n},mathrm{d}x
            &=int_0^{pi/2}tan^n(x/2)csc^2(x),mathrm{d}x\
            &=-int_0^{pi/2}tan^n(x/2),mathrm{d}cot(x)\
            &=-int_0^1u^n,mathrm{d}frac{1-u^2}{2u}\
            &=frac12int_0^1left(u^{n-2}+u^nright)mathrm{d}u\[3pt]
            &=frac12left(frac1{n-1}+frac1{n+1}right)\[6pt]
            &=frac{n}{n^2-1}
            end{align}
            $$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

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              2 Answers
              2






              active

              oldest

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              active

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              active

              oldest

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              7












              $begingroup$

              Note that the $csc^2(x)$ term can be expressed in terms of $tanleft(frac x2right)$ aswell. To be precise we got that



              begin{align*}
              csc^2(x)=left(frac1{sin(x)}right)^2=left(frac1{2sinleft(frac x2right)cosleft(frac x2right)}right)^2=left(frac{frac1{cos^2left(frac x2right)}}{2frac{sinleft(frac x2right)}{cosleft(frac x2right)}}right)^2
              =left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2
              end{align*}



              Using this and further noticing that $frac{mathrm d}{mathrm dx}tanleft(frac x2right)=frac12left(1+tan^2left(frac x2right)right)$ we may enforce the substition $tanleft(frac x2right)=u$ to obtain



              begin{align*}
              I_n=int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx&=int_0^{pi/2}tan^nleft(frac x2right)left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2mathrm dx\
              &=frac12int_0^{pi/2}tan^{n-2}left(frac x2right)left(1+tan^2left(frac x2right)right)left[frac12left(1+tan^2left(frac x2right)right)mathrm dxright]\
              &=frac12int_0^1 u^{n-2}(1+u^2)mathrm du\
              &=frac12left[frac{u^{n-1}}{n-1}+frac{u^{n+1}}{n+1}right]_0^1\
              &=frac12left[frac1{n-1}+frac1{n+1}right]
              end{align*}




              $$therefore~I_n~=~int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx~=~frac n{n^2-1}$$







              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                Note that the $csc^2(x)$ term can be expressed in terms of $tanleft(frac x2right)$ aswell. To be precise we got that



                begin{align*}
                csc^2(x)=left(frac1{sin(x)}right)^2=left(frac1{2sinleft(frac x2right)cosleft(frac x2right)}right)^2=left(frac{frac1{cos^2left(frac x2right)}}{2frac{sinleft(frac x2right)}{cosleft(frac x2right)}}right)^2
                =left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2
                end{align*}



                Using this and further noticing that $frac{mathrm d}{mathrm dx}tanleft(frac x2right)=frac12left(1+tan^2left(frac x2right)right)$ we may enforce the substition $tanleft(frac x2right)=u$ to obtain



                begin{align*}
                I_n=int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx&=int_0^{pi/2}tan^nleft(frac x2right)left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2mathrm dx\
                &=frac12int_0^{pi/2}tan^{n-2}left(frac x2right)left(1+tan^2left(frac x2right)right)left[frac12left(1+tan^2left(frac x2right)right)mathrm dxright]\
                &=frac12int_0^1 u^{n-2}(1+u^2)mathrm du\
                &=frac12left[frac{u^{n-1}}{n-1}+frac{u^{n+1}}{n+1}right]_0^1\
                &=frac12left[frac1{n-1}+frac1{n+1}right]
                end{align*}




                $$therefore~I_n~=~int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx~=~frac n{n^2-1}$$







                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Note that the $csc^2(x)$ term can be expressed in terms of $tanleft(frac x2right)$ aswell. To be precise we got that



                  begin{align*}
                  csc^2(x)=left(frac1{sin(x)}right)^2=left(frac1{2sinleft(frac x2right)cosleft(frac x2right)}right)^2=left(frac{frac1{cos^2left(frac x2right)}}{2frac{sinleft(frac x2right)}{cosleft(frac x2right)}}right)^2
                  =left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2
                  end{align*}



                  Using this and further noticing that $frac{mathrm d}{mathrm dx}tanleft(frac x2right)=frac12left(1+tan^2left(frac x2right)right)$ we may enforce the substition $tanleft(frac x2right)=u$ to obtain



                  begin{align*}
                  I_n=int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx&=int_0^{pi/2}tan^nleft(frac x2right)left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2mathrm dx\
                  &=frac12int_0^{pi/2}tan^{n-2}left(frac x2right)left(1+tan^2left(frac x2right)right)left[frac12left(1+tan^2left(frac x2right)right)mathrm dxright]\
                  &=frac12int_0^1 u^{n-2}(1+u^2)mathrm du\
                  &=frac12left[frac{u^{n-1}}{n-1}+frac{u^{n+1}}{n+1}right]_0^1\
                  &=frac12left[frac1{n-1}+frac1{n+1}right]
                  end{align*}




                  $$therefore~I_n~=~int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx~=~frac n{n^2-1}$$







                  share|cite|improve this answer









                  $endgroup$



                  Note that the $csc^2(x)$ term can be expressed in terms of $tanleft(frac x2right)$ aswell. To be precise we got that



                  begin{align*}
                  csc^2(x)=left(frac1{sin(x)}right)^2=left(frac1{2sinleft(frac x2right)cosleft(frac x2right)}right)^2=left(frac{frac1{cos^2left(frac x2right)}}{2frac{sinleft(frac x2right)}{cosleft(frac x2right)}}right)^2
                  =left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2
                  end{align*}



                  Using this and further noticing that $frac{mathrm d}{mathrm dx}tanleft(frac x2right)=frac12left(1+tan^2left(frac x2right)right)$ we may enforce the substition $tanleft(frac x2right)=u$ to obtain



                  begin{align*}
                  I_n=int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx&=int_0^{pi/2}tan^nleft(frac x2right)left(frac{1+tan^2left(frac x2right)}{2tanleft(frac x2right)}right)^2mathrm dx\
                  &=frac12int_0^{pi/2}tan^{n-2}left(frac x2right)left(1+tan^2left(frac x2right)right)left[frac12left(1+tan^2left(frac x2right)right)mathrm dxright]\
                  &=frac12int_0^1 u^{n-2}(1+u^2)mathrm du\
                  &=frac12left[frac{u^{n-1}}{n-1}+frac{u^{n+1}}{n+1}right]_0^1\
                  &=frac12left[frac1{n-1}+frac1{n+1}right]
                  end{align*}




                  $$therefore~I_n~=~int_0^{pi/2}tan^nleft(frac x2right)csc^2(x)mathrm dx~=~frac n{n^2-1}$$








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 5 hours ago









                  mrtaurhomrtaurho

                  4,85641235




                  4,85641235























                      5












                      $begingroup$

                      Slightly different approach, same answer.



                      Let $u=tan(x/2)$, then
                      $$
                      begin{align}
                      int_0^{pi/2}frac{sin^{n-2}(x)}{(1+cos(x))^n},mathrm{d}x
                      &=int_0^{pi/2}tan^n(x/2)csc^2(x),mathrm{d}x\
                      &=-int_0^{pi/2}tan^n(x/2),mathrm{d}cot(x)\
                      &=-int_0^1u^n,mathrm{d}frac{1-u^2}{2u}\
                      &=frac12int_0^1left(u^{n-2}+u^nright)mathrm{d}u\[3pt]
                      &=frac12left(frac1{n-1}+frac1{n+1}right)\[6pt]
                      &=frac{n}{n^2-1}
                      end{align}
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        5












                        $begingroup$

                        Slightly different approach, same answer.



                        Let $u=tan(x/2)$, then
                        $$
                        begin{align}
                        int_0^{pi/2}frac{sin^{n-2}(x)}{(1+cos(x))^n},mathrm{d}x
                        &=int_0^{pi/2}tan^n(x/2)csc^2(x),mathrm{d}x\
                        &=-int_0^{pi/2}tan^n(x/2),mathrm{d}cot(x)\
                        &=-int_0^1u^n,mathrm{d}frac{1-u^2}{2u}\
                        &=frac12int_0^1left(u^{n-2}+u^nright)mathrm{d}u\[3pt]
                        &=frac12left(frac1{n-1}+frac1{n+1}right)\[6pt]
                        &=frac{n}{n^2-1}
                        end{align}
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          5












                          5








                          5





                          $begingroup$

                          Slightly different approach, same answer.



                          Let $u=tan(x/2)$, then
                          $$
                          begin{align}
                          int_0^{pi/2}frac{sin^{n-2}(x)}{(1+cos(x))^n},mathrm{d}x
                          &=int_0^{pi/2}tan^n(x/2)csc^2(x),mathrm{d}x\
                          &=-int_0^{pi/2}tan^n(x/2),mathrm{d}cot(x)\
                          &=-int_0^1u^n,mathrm{d}frac{1-u^2}{2u}\
                          &=frac12int_0^1left(u^{n-2}+u^nright)mathrm{d}u\[3pt]
                          &=frac12left(frac1{n-1}+frac1{n+1}right)\[6pt]
                          &=frac{n}{n^2-1}
                          end{align}
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          Slightly different approach, same answer.



                          Let $u=tan(x/2)$, then
                          $$
                          begin{align}
                          int_0^{pi/2}frac{sin^{n-2}(x)}{(1+cos(x))^n},mathrm{d}x
                          &=int_0^{pi/2}tan^n(x/2)csc^2(x),mathrm{d}x\
                          &=-int_0^{pi/2}tan^n(x/2),mathrm{d}cot(x)\
                          &=-int_0^1u^n,mathrm{d}frac{1-u^2}{2u}\
                          &=frac12int_0^1left(u^{n-2}+u^nright)mathrm{d}u\[3pt]
                          &=frac12left(frac1{n-1}+frac1{n+1}right)\[6pt]
                          &=frac{n}{n^2-1}
                          end{align}
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 4 hours ago









                          robjohn♦robjohn

                          267k27308632




                          267k27308632






























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