Solving $int frac1{cos x}mathrm dx$ [duplicate]












2












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This question already has an answer here:




  • How to integrate $int frac{1}{cos(x)},mathrm dx$

    7 answers




I'm trying to solve the following integral:




$$int frac1{cos x}mathrm dx$$




I know that $int cos xmathrm dx$ = sin x, but I don't know how to proceed with $frac1{cos x}$.






calculus indefinite-integrals







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marked as duplicate by amWhy, Hans Lundmark, Abcd, metamorphy, Community♦ Jan 5 at 18:41


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  • 3




    $begingroup$
    Hint: multiply by $frac{cos x}{cos x}$ and use that $cos^2x=1-sin^2x$. Then perform a substitution to turn it into an integral of a rational function.
    $endgroup$
    – greelious
    Jan 5 at 16:43


















2












$begingroup$



This question already has an answer here:




  • How to integrate $int frac{1}{cos(x)},mathrm dx$

    7 answers




I'm trying to solve the following integral:




$$int frac1{cos x}mathrm dx$$




I know that $int cos xmathrm dx$ = sin x, but I don't know how to proceed with $frac1{cos x}$.






calculus indefinite-integrals







share|cite|improve this question











$endgroup$



marked as duplicate by amWhy, Hans Lundmark, Abcd, metamorphy, Community♦ Jan 5 at 18:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 3




    $begingroup$
    Hint: multiply by $frac{cos x}{cos x}$ and use that $cos^2x=1-sin^2x$. Then perform a substitution to turn it into an integral of a rational function.
    $endgroup$
    – greelious
    Jan 5 at 16:43
















2












2








2





$begingroup$



This question already has an answer here:




  • How to integrate $int frac{1}{cos(x)},mathrm dx$

    7 answers




I'm trying to solve the following integral:




$$int frac1{cos x}mathrm dx$$




I know that $int cos xmathrm dx$ = sin x, but I don't know how to proceed with $frac1{cos x}$.






calculus indefinite-integrals







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • How to integrate $int frac{1}{cos(x)},mathrm dx$

    7 answers




I'm trying to solve the following integral:




$$int frac1{cos x}mathrm dx$$




I know that $int cos xmathrm dx$ = sin x, but I don't know how to proceed with $frac1{cos x}$.





This question already has an answer here:




  • How to integrate $int frac{1}{cos(x)},mathrm dx$

    7 answers







calculus indefinite-integrals




calculus indefinite-integrals






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edited Jan 5 at 16:44









mrtaurho

4,13621234




4,13621234










asked Jan 5 at 16:39









El BryanEl Bryan

417




417




marked as duplicate by amWhy, Hans Lundmark, Abcd, metamorphy, Community♦ Jan 5 at 18:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by amWhy, Hans Lundmark, Abcd, metamorphy, Community♦ Jan 5 at 18:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 3




    $begingroup$
    Hint: multiply by $frac{cos x}{cos x}$ and use that $cos^2x=1-sin^2x$. Then perform a substitution to turn it into an integral of a rational function.
    $endgroup$
    – greelious
    Jan 5 at 16:43
















  • 3




    $begingroup$
    Hint: multiply by $frac{cos x}{cos x}$ and use that $cos^2x=1-sin^2x$. Then perform a substitution to turn it into an integral of a rational function.
    $endgroup$
    – greelious
    Jan 5 at 16:43










3




3




$begingroup$
Hint: multiply by $frac{cos x}{cos x}$ and use that $cos^2x=1-sin^2x$. Then perform a substitution to turn it into an integral of a rational function.
$endgroup$
– greelious
Jan 5 at 16:43






$begingroup$
Hint: multiply by $frac{cos x}{cos x}$ and use that $cos^2x=1-sin^2x$. Then perform a substitution to turn it into an integral of a rational function.
$endgroup$
– greelious
Jan 5 at 16:43












4 Answers
4






active

oldest

votes


















5












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$$intfrac{dx}{cos x}=intfrac{cos x}{1-sin^2x}dx=frac12intcos xleft(frac1{1-sin x}+frac1{1+sin x}right),dx=$$



$$frac12left[-intfrac{d(1-sin x)}{1-sin x}+intfrac{d(1+sin x)}{1+sin x}right]=frac12logfrac{1+sin x}{1-sin x}+C$$






share|cite|improve this answer









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    5












    $begingroup$

    $$int sec(x)dx$$Let $u=tan(frac{x}{2})$
    $$int sec(x)dx=2intfrac{du}{1-u^2}$$ Use partial fractions to get $$int sec(x)dx=intfrac{du}{1-u}-intfrac{du}{1+u}=lnleft(frac{1-u}{1+u}right)+C=lnleft(frac{1-tan(frac{x}{2})}{1+tan(frac{x}{2})}right)+C$$






    share|cite|improve this answer











    $endgroup$





















      5












      $begingroup$

      Utilizing the given hint from greelious we get the following



      $$I=int frac{mathrm dx}{cos x}=int frac{cos x}{cos^2 x}mathrm dx=int frac{cos x}{1-sin^2 x}mathrm dx$$



      Now by the substitution $y=sin x$ we further get



      $$I=int frac{cos x}{1-sin^2 x}mathrm dxstackrel{y=sin x}{=}int frac{mathrm dy}{1-y^2}$$



      This on can be done via partial fraction decomposition which gives us



      $$I=int frac{mathrm dy}{1-y^2}=frac12int left(frac1{1+y}+frac1{1-y}right)mathrm dy=frac12logleft(frac{1+y}{1-y}right)+c$$



      And now resubstitute $y=sin(x)$ gives us



      $$I=frac12logleft(frac{1+y}{1-y}right)+c=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




      $$therefore~I=intfrac{mathrm dx}{cos x}=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




      Moreover we can show that this solution equals the elegant one given by Rhys Hughes



      $$begin{align*}
      &color{red}{frac12logleft(frac{1+sin x}{1-sin x}right)}=logleft(sqrt{frac{1+cos left(x-fracpi2right)}{1-cosleft(x-fracpi2right)}}right)=logleft(cotleft(frac{x-fracpi2}{2}right)right)\&=logleft(cotleft(frac x2-fracpi4right)right)
      =logleft(frac{1+cotfrac x2}{1-cot frac x2}right)=logleft(frac{left(cosfrac x2+sinfrac x2right)^2}{cos^2 frac x2-sin^2frac x2}right)\&=logleft(frac{1+2sinfrac x2cosfrac x2}{cos x}right)=logleft(frac{1+sin x}{cos x}right)=color{red}{log(sec x+tan x)}
      end{align*}$$






      share|cite|improve this answer











      $endgroup$





















        4












        $begingroup$

        A method I've seen is:



        $$int sec(x)frac{sec x+tan x}{sec x+tan x}dx=intfrac{(sec x+tan x)'}{sec x+tan x}dx=ln|sec x+tan x|+C$$






        share|cite|improve this answer











        $endgroup$









        • 2




          $begingroup$
          Whoops. Added, cheers.
          $endgroup$
          – Rhys Hughes
          Jan 5 at 18:29


















        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        $$intfrac{dx}{cos x}=intfrac{cos x}{1-sin^2x}dx=frac12intcos xleft(frac1{1-sin x}+frac1{1+sin x}right),dx=$$



        $$frac12left[-intfrac{d(1-sin x)}{1-sin x}+intfrac{d(1+sin x)}{1+sin x}right]=frac12logfrac{1+sin x}{1-sin x}+C$$






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          $$intfrac{dx}{cos x}=intfrac{cos x}{1-sin^2x}dx=frac12intcos xleft(frac1{1-sin x}+frac1{1+sin x}right),dx=$$



          $$frac12left[-intfrac{d(1-sin x)}{1-sin x}+intfrac{d(1+sin x)}{1+sin x}right]=frac12logfrac{1+sin x}{1-sin x}+C$$






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            $$intfrac{dx}{cos x}=intfrac{cos x}{1-sin^2x}dx=frac12intcos xleft(frac1{1-sin x}+frac1{1+sin x}right),dx=$$



            $$frac12left[-intfrac{d(1-sin x)}{1-sin x}+intfrac{d(1+sin x)}{1+sin x}right]=frac12logfrac{1+sin x}{1-sin x}+C$$






            share|cite|improve this answer









            $endgroup$



            $$intfrac{dx}{cos x}=intfrac{cos x}{1-sin^2x}dx=frac12intcos xleft(frac1{1-sin x}+frac1{1+sin x}right),dx=$$



            $$frac12left[-intfrac{d(1-sin x)}{1-sin x}+intfrac{d(1+sin x)}{1+sin x}right]=frac12logfrac{1+sin x}{1-sin x}+C$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 16:49









            DonAntonioDonAntonio

            178k1492227




            178k1492227























                5












                $begingroup$

                $$int sec(x)dx$$Let $u=tan(frac{x}{2})$
                $$int sec(x)dx=2intfrac{du}{1-u^2}$$ Use partial fractions to get $$int sec(x)dx=intfrac{du}{1-u}-intfrac{du}{1+u}=lnleft(frac{1-u}{1+u}right)+C=lnleft(frac{1-tan(frac{x}{2})}{1+tan(frac{x}{2})}right)+C$$






                share|cite|improve this answer











                $endgroup$


















                  5












                  $begingroup$

                  $$int sec(x)dx$$Let $u=tan(frac{x}{2})$
                  $$int sec(x)dx=2intfrac{du}{1-u^2}$$ Use partial fractions to get $$int sec(x)dx=intfrac{du}{1-u}-intfrac{du}{1+u}=lnleft(frac{1-u}{1+u}right)+C=lnleft(frac{1-tan(frac{x}{2})}{1+tan(frac{x}{2})}right)+C$$






                  share|cite|improve this answer











                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    $$int sec(x)dx$$Let $u=tan(frac{x}{2})$
                    $$int sec(x)dx=2intfrac{du}{1-u^2}$$ Use partial fractions to get $$int sec(x)dx=intfrac{du}{1-u}-intfrac{du}{1+u}=lnleft(frac{1-u}{1+u}right)+C=lnleft(frac{1-tan(frac{x}{2})}{1+tan(frac{x}{2})}right)+C$$






                    share|cite|improve this answer











                    $endgroup$



                    $$int sec(x)dx$$Let $u=tan(frac{x}{2})$
                    $$int sec(x)dx=2intfrac{du}{1-u^2}$$ Use partial fractions to get $$int sec(x)dx=intfrac{du}{1-u}-intfrac{du}{1+u}=lnleft(frac{1-u}{1+u}right)+C=lnleft(frac{1-tan(frac{x}{2})}{1+tan(frac{x}{2})}right)+C$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 5 at 17:26

























                    answered Jan 5 at 16:46









                    aledenaleden

                    2,037511




                    2,037511























                        5












                        $begingroup$

                        Utilizing the given hint from greelious we get the following



                        $$I=int frac{mathrm dx}{cos x}=int frac{cos x}{cos^2 x}mathrm dx=int frac{cos x}{1-sin^2 x}mathrm dx$$



                        Now by the substitution $y=sin x$ we further get



                        $$I=int frac{cos x}{1-sin^2 x}mathrm dxstackrel{y=sin x}{=}int frac{mathrm dy}{1-y^2}$$



                        This on can be done via partial fraction decomposition which gives us



                        $$I=int frac{mathrm dy}{1-y^2}=frac12int left(frac1{1+y}+frac1{1-y}right)mathrm dy=frac12logleft(frac{1+y}{1-y}right)+c$$



                        And now resubstitute $y=sin(x)$ gives us



                        $$I=frac12logleft(frac{1+y}{1-y}right)+c=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                        $$therefore~I=intfrac{mathrm dx}{cos x}=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                        Moreover we can show that this solution equals the elegant one given by Rhys Hughes



                        $$begin{align*}
                        &color{red}{frac12logleft(frac{1+sin x}{1-sin x}right)}=logleft(sqrt{frac{1+cos left(x-fracpi2right)}{1-cosleft(x-fracpi2right)}}right)=logleft(cotleft(frac{x-fracpi2}{2}right)right)\&=logleft(cotleft(frac x2-fracpi4right)right)
                        =logleft(frac{1+cotfrac x2}{1-cot frac x2}right)=logleft(frac{left(cosfrac x2+sinfrac x2right)^2}{cos^2 frac x2-sin^2frac x2}right)\&=logleft(frac{1+2sinfrac x2cosfrac x2}{cos x}right)=logleft(frac{1+sin x}{cos x}right)=color{red}{log(sec x+tan x)}
                        end{align*}$$






                        share|cite|improve this answer











                        $endgroup$


















                          5












                          $begingroup$

                          Utilizing the given hint from greelious we get the following



                          $$I=int frac{mathrm dx}{cos x}=int frac{cos x}{cos^2 x}mathrm dx=int frac{cos x}{1-sin^2 x}mathrm dx$$



                          Now by the substitution $y=sin x$ we further get



                          $$I=int frac{cos x}{1-sin^2 x}mathrm dxstackrel{y=sin x}{=}int frac{mathrm dy}{1-y^2}$$



                          This on can be done via partial fraction decomposition which gives us



                          $$I=int frac{mathrm dy}{1-y^2}=frac12int left(frac1{1+y}+frac1{1-y}right)mathrm dy=frac12logleft(frac{1+y}{1-y}right)+c$$



                          And now resubstitute $y=sin(x)$ gives us



                          $$I=frac12logleft(frac{1+y}{1-y}right)+c=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                          $$therefore~I=intfrac{mathrm dx}{cos x}=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                          Moreover we can show that this solution equals the elegant one given by Rhys Hughes



                          $$begin{align*}
                          &color{red}{frac12logleft(frac{1+sin x}{1-sin x}right)}=logleft(sqrt{frac{1+cos left(x-fracpi2right)}{1-cosleft(x-fracpi2right)}}right)=logleft(cotleft(frac{x-fracpi2}{2}right)right)\&=logleft(cotleft(frac x2-fracpi4right)right)
                          =logleft(frac{1+cotfrac x2}{1-cot frac x2}right)=logleft(frac{left(cosfrac x2+sinfrac x2right)^2}{cos^2 frac x2-sin^2frac x2}right)\&=logleft(frac{1+2sinfrac x2cosfrac x2}{cos x}right)=logleft(frac{1+sin x}{cos x}right)=color{red}{log(sec x+tan x)}
                          end{align*}$$






                          share|cite|improve this answer











                          $endgroup$
















                            5












                            5








                            5





                            $begingroup$

                            Utilizing the given hint from greelious we get the following



                            $$I=int frac{mathrm dx}{cos x}=int frac{cos x}{cos^2 x}mathrm dx=int frac{cos x}{1-sin^2 x}mathrm dx$$



                            Now by the substitution $y=sin x$ we further get



                            $$I=int frac{cos x}{1-sin^2 x}mathrm dxstackrel{y=sin x}{=}int frac{mathrm dy}{1-y^2}$$



                            This on can be done via partial fraction decomposition which gives us



                            $$I=int frac{mathrm dy}{1-y^2}=frac12int left(frac1{1+y}+frac1{1-y}right)mathrm dy=frac12logleft(frac{1+y}{1-y}right)+c$$



                            And now resubstitute $y=sin(x)$ gives us



                            $$I=frac12logleft(frac{1+y}{1-y}right)+c=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                            $$therefore~I=intfrac{mathrm dx}{cos x}=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                            Moreover we can show that this solution equals the elegant one given by Rhys Hughes



                            $$begin{align*}
                            &color{red}{frac12logleft(frac{1+sin x}{1-sin x}right)}=logleft(sqrt{frac{1+cos left(x-fracpi2right)}{1-cosleft(x-fracpi2right)}}right)=logleft(cotleft(frac{x-fracpi2}{2}right)right)\&=logleft(cotleft(frac x2-fracpi4right)right)
                            =logleft(frac{1+cotfrac x2}{1-cot frac x2}right)=logleft(frac{left(cosfrac x2+sinfrac x2right)^2}{cos^2 frac x2-sin^2frac x2}right)\&=logleft(frac{1+2sinfrac x2cosfrac x2}{cos x}right)=logleft(frac{1+sin x}{cos x}right)=color{red}{log(sec x+tan x)}
                            end{align*}$$






                            share|cite|improve this answer











                            $endgroup$



                            Utilizing the given hint from greelious we get the following



                            $$I=int frac{mathrm dx}{cos x}=int frac{cos x}{cos^2 x}mathrm dx=int frac{cos x}{1-sin^2 x}mathrm dx$$



                            Now by the substitution $y=sin x$ we further get



                            $$I=int frac{cos x}{1-sin^2 x}mathrm dxstackrel{y=sin x}{=}int frac{mathrm dy}{1-y^2}$$



                            This on can be done via partial fraction decomposition which gives us



                            $$I=int frac{mathrm dy}{1-y^2}=frac12int left(frac1{1+y}+frac1{1-y}right)mathrm dy=frac12logleft(frac{1+y}{1-y}right)+c$$



                            And now resubstitute $y=sin(x)$ gives us



                            $$I=frac12logleft(frac{1+y}{1-y}right)+c=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                            $$therefore~I=intfrac{mathrm dx}{cos x}=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$




                            Moreover we can show that this solution equals the elegant one given by Rhys Hughes



                            $$begin{align*}
                            &color{red}{frac12logleft(frac{1+sin x}{1-sin x}right)}=logleft(sqrt{frac{1+cos left(x-fracpi2right)}{1-cosleft(x-fracpi2right)}}right)=logleft(cotleft(frac{x-fracpi2}{2}right)right)\&=logleft(cotleft(frac x2-fracpi4right)right)
                            =logleft(frac{1+cotfrac x2}{1-cot frac x2}right)=logleft(frac{left(cosfrac x2+sinfrac x2right)^2}{cos^2 frac x2-sin^2frac x2}right)\&=logleft(frac{1+2sinfrac x2cosfrac x2}{cos x}right)=logleft(frac{1+sin x}{cos x}right)=color{red}{log(sec x+tan x)}
                            end{align*}$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 9 at 9:33

























                            answered Jan 5 at 16:54









                            mrtaurhomrtaurho

                            4,13621234




                            4,13621234























                                4












                                $begingroup$

                                A method I've seen is:



                                $$int sec(x)frac{sec x+tan x}{sec x+tan x}dx=intfrac{(sec x+tan x)'}{sec x+tan x}dx=ln|sec x+tan x|+C$$






                                share|cite|improve this answer











                                $endgroup$









                                • 2




                                  $begingroup$
                                  Whoops. Added, cheers.
                                  $endgroup$
                                  – Rhys Hughes
                                  Jan 5 at 18:29
















                                4












                                $begingroup$

                                A method I've seen is:



                                $$int sec(x)frac{sec x+tan x}{sec x+tan x}dx=intfrac{(sec x+tan x)'}{sec x+tan x}dx=ln|sec x+tan x|+C$$






                                share|cite|improve this answer











                                $endgroup$









                                • 2




                                  $begingroup$
                                  Whoops. Added, cheers.
                                  $endgroup$
                                  – Rhys Hughes
                                  Jan 5 at 18:29














                                4












                                4








                                4





                                $begingroup$

                                A method I've seen is:



                                $$int sec(x)frac{sec x+tan x}{sec x+tan x}dx=intfrac{(sec x+tan x)'}{sec x+tan x}dx=ln|sec x+tan x|+C$$






                                share|cite|improve this answer











                                $endgroup$



                                A method I've seen is:



                                $$int sec(x)frac{sec x+tan x}{sec x+tan x}dx=intfrac{(sec x+tan x)'}{sec x+tan x}dx=ln|sec x+tan x|+C$$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jan 5 at 18:28

























                                answered Jan 5 at 16:50









                                Rhys HughesRhys Hughes

                                5,9791529




                                5,9791529








                                • 2




                                  $begingroup$
                                  Whoops. Added, cheers.
                                  $endgroup$
                                  – Rhys Hughes
                                  Jan 5 at 18:29














                                • 2




                                  $begingroup$
                                  Whoops. Added, cheers.
                                  $endgroup$
                                  – Rhys Hughes
                                  Jan 5 at 18:29








                                2




                                2




                                $begingroup$
                                Whoops. Added, cheers.
                                $endgroup$
                                – Rhys Hughes
                                Jan 5 at 18:29




                                $begingroup$
                                Whoops. Added, cheers.
                                $endgroup$
                                – Rhys Hughes
                                Jan 5 at 18:29



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