Integral $int frac{sin^n(x)}{cos(x)}dx$












3














In one of my exercises about integration we had to solve the following integral:



begin{equation}
int frac{sin^n(x)}{cos^m(x)}dx
end{equation}



We had to do this via a recursive integral. I found:



begin{equation}
mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
end{equation}



I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.



begin{equation}
int frac{sin^n(x)}{cos(x)}dx
end{equation}



I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:





  • $u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$


  • $u = tan(x)$, $v = sin^{n-1}(x)$


  • $u = sin^{n-1}(x)$, $v = tan(x)$


Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.










share|cite|improve this question





























    3














    In one of my exercises about integration we had to solve the following integral:



    begin{equation}
    int frac{sin^n(x)}{cos^m(x)}dx
    end{equation}



    We had to do this via a recursive integral. I found:



    begin{equation}
    mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
    end{equation}



    I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.



    begin{equation}
    int frac{sin^n(x)}{cos(x)}dx
    end{equation}



    I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:





    • $u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$


    • $u = tan(x)$, $v = sin^{n-1}(x)$


    • $u = sin^{n-1}(x)$, $v = tan(x)$


    Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.










    share|cite|improve this question



























      3












      3








      3


      1





      In one of my exercises about integration we had to solve the following integral:



      begin{equation}
      int frac{sin^n(x)}{cos^m(x)}dx
      end{equation}



      We had to do this via a recursive integral. I found:



      begin{equation}
      mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
      end{equation}



      I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.



      begin{equation}
      int frac{sin^n(x)}{cos(x)}dx
      end{equation}



      I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:





      • $u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$


      • $u = tan(x)$, $v = sin^{n-1}(x)$


      • $u = sin^{n-1}(x)$, $v = tan(x)$


      Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.










      share|cite|improve this question















      In one of my exercises about integration we had to solve the following integral:



      begin{equation}
      int frac{sin^n(x)}{cos^m(x)}dx
      end{equation}



      We had to do this via a recursive integral. I found:



      begin{equation}
      mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
      end{equation}



      I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.



      begin{equation}
      int frac{sin^n(x)}{cos(x)}dx
      end{equation}



      I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:





      • $u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$


      • $u = tan(x)$, $v = sin^{n-1}(x)$


      • $u = sin^{n-1}(x)$, $v = tan(x)$


      Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.







      integration indefinite-integrals recursion reduction-formula






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      edited Dec 27 '18 at 23:48









      clathratus

      3,243331




      3,243331










      asked Dec 27 '18 at 14:30









      Viktor

      416




      416






















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          Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
          begin{align*}
          I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
          &=int sin^n(x)cos(x)~mathrm dx,
          end{align*}

          which is the reduction formula for $I_n$.






          share|cite|improve this answer





















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            4














            Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
            begin{align*}
            I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
            &=int sin^n(x)cos(x)~mathrm dx,
            end{align*}

            which is the reduction formula for $I_n$.






            share|cite|improve this answer


























              4














              Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
              begin{align*}
              I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
              &=int sin^n(x)cos(x)~mathrm dx,
              end{align*}

              which is the reduction formula for $I_n$.






              share|cite|improve this answer
























                4












                4








                4






                Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
                begin{align*}
                I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
                &=int sin^n(x)cos(x)~mathrm dx,
                end{align*}

                which is the reduction formula for $I_n$.






                share|cite|improve this answer












                Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
                begin{align*}
                I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
                &=int sin^n(x)cos(x)~mathrm dx,
                end{align*}

                which is the reduction formula for $I_n$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 27 '18 at 14:41









                Tianlalu

                3,09621038




                3,09621038






























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