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












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?






integration definite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$

















    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?






    integration 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?






      integration 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?






      integration definite-integrals trigonometric-integrals




      integration 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 Feb 11 at 6:41









      user21820

      39k543153




      39k543153










      asked Feb 10 at 18:43









      jackyjacky

      856612




      856612






















          2 Answers
          2






          active

          oldest

          votes


















          8












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            For anyone following along - pay attention to where the limits of the integration change along with the change of variable from $dx$ to $du$.
            $endgroup$
            – Stobor
            Feb 11 at 2:12



















          6












          $begingroup$

          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

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              For anyone following along - pay attention to where the limits of the integration change along with the change of variable from $dx$ to $du$.
              $endgroup$
              – Stobor
              Feb 11 at 2:12
















            8












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              For anyone following along - pay attention to where the limits of the integration change along with the change of variable from $dx$ to $du$.
              $endgroup$
              – Stobor
              Feb 11 at 2:12














            8












            8








            8





            $begingroup$

            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 Feb 10 at 19:04









            mrtaurhomrtaurho

            5,46541237




            5,46541237












            • $begingroup$
              For anyone following along - pay attention to where the limits of the integration change along with the change of variable from $dx$ to $du$.
              $endgroup$
              – Stobor
              Feb 11 at 2:12


















            • $begingroup$
              For anyone following along - pay attention to where the limits of the integration change along with the change of variable from $dx$ to $du$.
              $endgroup$
              – Stobor
              Feb 11 at 2:12
















            $begingroup$
            For anyone following along - pay attention to where the limits of the integration change along with the change of variable from $dx$ to $du$.
            $endgroup$
            – Stobor
            Feb 11 at 2:12




            $begingroup$
            For anyone following along - pay attention to where the limits of the integration change along with the change of variable from $dx$ to $du$.
            $endgroup$
            – Stobor
            Feb 11 at 2:12











            6












            $begingroup$

            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









            $endgroup$


















              6












              $begingroup$

              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









              $endgroup$
















                6












                6








                6





                $begingroup$

                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 Feb 10 at 20:14









                robjohn♦robjohn

                268k27308633




                268k27308633






























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