Integration by Substitution of $cos(log(x))$












0












$begingroup$


I need to find the indefinite integral:
$int cos(log(x)) dx$.



I've tried using u-substitution, setting $u=log(x)$. Then I get:
$int cos(u)du$, where $du$ is $frac{1}{x}dx$.



The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?



Thanks for your time.






integration substitution







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I need to find the indefinite integral:
    $int cos(log(x)) dx$.



    I've tried using u-substitution, setting $u=log(x)$. Then I get:
    $int cos(u)du$, where $du$ is $frac{1}{x}dx$.



    The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?



    Thanks for your time.






    integration substitution







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I need to find the indefinite integral:
      $int cos(log(x)) dx$.



      I've tried using u-substitution, setting $u=log(x)$. Then I get:
      $int cos(u)du$, where $du$ is $frac{1}{x}dx$.



      The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?



      Thanks for your time.






      integration substitution







      share|cite|improve this question











      $endgroup$




      I need to find the indefinite integral:
      $int cos(log(x)) dx$.



      I've tried using u-substitution, setting $u=log(x)$. Then I get:
      $int cos(u)du$, where $du$ is $frac{1}{x}dx$.



      The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?



      Thanks for your time.






      integration substitution




      integration substitution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited May 13 '18 at 7:56









      farruhota

      20.4k2739




      20.4k2739










      asked May 13 '18 at 7:27









      Three OneFourThree OneFour

      1099




      1099






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Let $ln x=uimplies x=e^u,dx=e^u du$



          $$intcos(ln x)dx=int e^ucos u du $$



          Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$



          See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How do you know that $dx=e^u du$?
            $endgroup$
            – Three OneFour
            May 13 '18 at 7:34










          • $begingroup$
            @ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
            $endgroup$
            – lab bhattacharjee
            May 13 '18 at 7:35



















          0












          $begingroup$

          Better way:



          Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
          So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
          And hence
          $$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
          Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
          We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Let $ln x=uimplies x=e^u,dx=e^u du$



            $$intcos(ln x)dx=int e^ucos u du $$



            Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$



            See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How do you know that $dx=e^u du$?
              $endgroup$
              – Three OneFour
              May 13 '18 at 7:34










            • $begingroup$
              @ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
              $endgroup$
              – lab bhattacharjee
              May 13 '18 at 7:35
















            2












            $begingroup$

            Let $ln x=uimplies x=e^u,dx=e^u du$



            $$intcos(ln x)dx=int e^ucos u du $$



            Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$



            See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How do you know that $dx=e^u du$?
              $endgroup$
              – Three OneFour
              May 13 '18 at 7:34










            • $begingroup$
              @ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
              $endgroup$
              – lab bhattacharjee
              May 13 '18 at 7:35














            2












            2








            2





            $begingroup$

            Let $ln x=uimplies x=e^u,dx=e^u du$



            $$intcos(ln x)dx=int e^ucos u du $$



            Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$



            See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$






            share|cite|improve this answer









            $endgroup$



            Let $ln x=uimplies x=e^u,dx=e^u du$



            $$intcos(ln x)dx=int e^ucos u du $$



            Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$



            See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 13 '18 at 7:32









            lab bhattacharjeelab bhattacharjee

            226k15157275




            226k15157275












            • $begingroup$
              How do you know that $dx=e^u du$?
              $endgroup$
              – Three OneFour
              May 13 '18 at 7:34










            • $begingroup$
              @ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
              $endgroup$
              – lab bhattacharjee
              May 13 '18 at 7:35


















            • $begingroup$
              How do you know that $dx=e^u du$?
              $endgroup$
              – Three OneFour
              May 13 '18 at 7:34










            • $begingroup$
              @ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
              $endgroup$
              – lab bhattacharjee
              May 13 '18 at 7:35
















            $begingroup$
            How do you know that $dx=e^u du$?
            $endgroup$
            – Three OneFour
            May 13 '18 at 7:34




            $begingroup$
            How do you know that $dx=e^u du$?
            $endgroup$
            – Three OneFour
            May 13 '18 at 7:34












            $begingroup$
            @ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
            $endgroup$
            – lab bhattacharjee
            May 13 '18 at 7:35




            $begingroup$
            @ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
            $endgroup$
            – lab bhattacharjee
            May 13 '18 at 7:35











            0












            $begingroup$

            Better way:



            Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
            So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
            And hence
            $$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
            Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
            We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Better way:



              Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
              So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
              And hence
              $$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
              Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
              We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Better way:



                Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
                So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
                And hence
                $$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
                Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
                We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$






                share|cite|improve this answer









                $endgroup$



                Better way:



                Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
                So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
                And hence
                $$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
                Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
                We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 1:42









                clathratusclathratus

                4,615337




                4,615337






























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