Can this integration be done by using substitution [on hold]












-1














Following function is given in calculus and should be solved by using substitution



$$int frac{sqrt{4x^2-1}}{x^4} dx$$



I tried to substitute each component but still did not get answer










share|cite|improve this question















put on hold as off-topic by RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist

If this question can be reworded to fit the rules in the help center, please edit the question.


















    -1














    Following function is given in calculus and should be solved by using substitution



    $$int frac{sqrt{4x^2-1}}{x^4} dx$$



    I tried to substitute each component but still did not get answer










    share|cite|improve this question















    put on hold as off-topic by RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist yesterday


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      -1












      -1








      -1







      Following function is given in calculus and should be solved by using substitution



      $$int frac{sqrt{4x^2-1}}{x^4} dx$$



      I tried to substitute each component but still did not get answer










      share|cite|improve this question















      Following function is given in calculus and should be solved by using substitution



      $$int frac{sqrt{4x^2-1}}{x^4} dx$$



      I tried to substitute each component but still did not get answer







      calculus linear-algebra integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Dylan

      12.2k31026




      12.2k31026










      asked Dec 23 at 20:42









      Arif Rustamov

      327




      327




      put on hold as off-topic by RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist

      If this question can be reworded to fit the rules in the help center, please edit the question.




      put on hold as off-topic by RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Davide Giraudo, Leucippus, Shailesh, ncmathsadist

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






          active

          oldest

          votes


















          5














          hint



          Put



          $$2x=cosh(t) text{ in } [frac 12,+infty)$$



          or
          $$2x=-cosh(t) text{ in } (-infty,-frac 12]$$
          to get



          $$sqrt{4x^2-1}=sqrt{cosh^2(t)-1}=sinh(t)$$






          share|cite|improve this answer































            5














            Hint. $displaystyleintfrac{sqrt{4x^2-1}}{x^4}dx=begin{cases}displaystyleintfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x>0\displaystyle-intfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x<0end{cases}$



            Keep $4-frac1{x^2}=timplies dt=2dx/x^3$.






            share|cite|improve this answer































              2














              This one is actually pretty neat
              $$I=intfrac{sqrt{4x^2-1}}{x^4}mathrm dx$$
              The first thing to do is get rid of the square root, so we do
              $$x=frac12sec uRightarrow mathrm dx=frac12sec utan u,mathrm du$$
              Hence
              $$I=frac12intfrac{sqrt{4cdotfrac14sec^2u-1}}{frac1{2^4}sec^4u}sec utan u,mathrm du$$
              Which simplifies to
              $$I=8intfrac{sqrt{sec^2u-1}}{sec^3u}tan u,mathrm du$$
              Noting that $sec^2u-1=tan^2u$, we can simplify more:
              $$I=8intfrac{tan^2u}{sec^3u}mathrm du$$
              Using the reciprocal and quotient identities this simplifies to
              $$I=8intsin^2ucos u,mathrm du$$
              Then we preform another substitution
              $$w=sin uRightarrow mathrm dw=cos u, mathrm du$$
              So we have
              $$I=8int w^2mathrm dw$$
              $$I=frac83w^3$$
              $$I=frac83sin^3u$$
              $$I=frac83(1-cos^2u)^{3/2}$$
              $$I=frac83bigg(1-frac1{sec^2u}bigg)^{3/2}$$
              since $x=frac12sec u$,
              $$I=frac83bigg(1-frac1{4x^2}bigg)^{3/2}$$
              $$I=frac{(4x^2-1)^{3/2}}{3x^3}+C$$






              share|cite|improve this answer




























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5














                hint



                Put



                $$2x=cosh(t) text{ in } [frac 12,+infty)$$



                or
                $$2x=-cosh(t) text{ in } (-infty,-frac 12]$$
                to get



                $$sqrt{4x^2-1}=sqrt{cosh^2(t)-1}=sinh(t)$$






                share|cite|improve this answer




























                  5














                  hint



                  Put



                  $$2x=cosh(t) text{ in } [frac 12,+infty)$$



                  or
                  $$2x=-cosh(t) text{ in } (-infty,-frac 12]$$
                  to get



                  $$sqrt{4x^2-1}=sqrt{cosh^2(t)-1}=sinh(t)$$






                  share|cite|improve this answer


























                    5












                    5








                    5






                    hint



                    Put



                    $$2x=cosh(t) text{ in } [frac 12,+infty)$$



                    or
                    $$2x=-cosh(t) text{ in } (-infty,-frac 12]$$
                    to get



                    $$sqrt{4x^2-1}=sqrt{cosh^2(t)-1}=sinh(t)$$






                    share|cite|improve this answer














                    hint



                    Put



                    $$2x=cosh(t) text{ in } [frac 12,+infty)$$



                    or
                    $$2x=-cosh(t) text{ in } (-infty,-frac 12]$$
                    to get



                    $$sqrt{4x^2-1}=sqrt{cosh^2(t)-1}=sinh(t)$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 23 at 20:51

























                    answered Dec 23 at 20:46









                    hamam_Abdallah

                    37.9k21634




                    37.9k21634























                        5














                        Hint. $displaystyleintfrac{sqrt{4x^2-1}}{x^4}dx=begin{cases}displaystyleintfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x>0\displaystyle-intfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x<0end{cases}$



                        Keep $4-frac1{x^2}=timplies dt=2dx/x^3$.






                        share|cite|improve this answer




























                          5














                          Hint. $displaystyleintfrac{sqrt{4x^2-1}}{x^4}dx=begin{cases}displaystyleintfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x>0\displaystyle-intfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x<0end{cases}$



                          Keep $4-frac1{x^2}=timplies dt=2dx/x^3$.






                          share|cite|improve this answer


























                            5












                            5








                            5






                            Hint. $displaystyleintfrac{sqrt{4x^2-1}}{x^4}dx=begin{cases}displaystyleintfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x>0\displaystyle-intfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x<0end{cases}$



                            Keep $4-frac1{x^2}=timplies dt=2dx/x^3$.






                            share|cite|improve this answer














                            Hint. $displaystyleintfrac{sqrt{4x^2-1}}{x^4}dx=begin{cases}displaystyleintfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x>0\displaystyle-intfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x<0end{cases}$



                            Keep $4-frac1{x^2}=timplies dt=2dx/x^3$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 23 at 21:10

























                            answered Dec 23 at 20:47









                            Shubham Johri

                            3,826716




                            3,826716























                                2














                                This one is actually pretty neat
                                $$I=intfrac{sqrt{4x^2-1}}{x^4}mathrm dx$$
                                The first thing to do is get rid of the square root, so we do
                                $$x=frac12sec uRightarrow mathrm dx=frac12sec utan u,mathrm du$$
                                Hence
                                $$I=frac12intfrac{sqrt{4cdotfrac14sec^2u-1}}{frac1{2^4}sec^4u}sec utan u,mathrm du$$
                                Which simplifies to
                                $$I=8intfrac{sqrt{sec^2u-1}}{sec^3u}tan u,mathrm du$$
                                Noting that $sec^2u-1=tan^2u$, we can simplify more:
                                $$I=8intfrac{tan^2u}{sec^3u}mathrm du$$
                                Using the reciprocal and quotient identities this simplifies to
                                $$I=8intsin^2ucos u,mathrm du$$
                                Then we preform another substitution
                                $$w=sin uRightarrow mathrm dw=cos u, mathrm du$$
                                So we have
                                $$I=8int w^2mathrm dw$$
                                $$I=frac83w^3$$
                                $$I=frac83sin^3u$$
                                $$I=frac83(1-cos^2u)^{3/2}$$
                                $$I=frac83bigg(1-frac1{sec^2u}bigg)^{3/2}$$
                                since $x=frac12sec u$,
                                $$I=frac83bigg(1-frac1{4x^2}bigg)^{3/2}$$
                                $$I=frac{(4x^2-1)^{3/2}}{3x^3}+C$$






                                share|cite|improve this answer


























                                  2














                                  This one is actually pretty neat
                                  $$I=intfrac{sqrt{4x^2-1}}{x^4}mathrm dx$$
                                  The first thing to do is get rid of the square root, so we do
                                  $$x=frac12sec uRightarrow mathrm dx=frac12sec utan u,mathrm du$$
                                  Hence
                                  $$I=frac12intfrac{sqrt{4cdotfrac14sec^2u-1}}{frac1{2^4}sec^4u}sec utan u,mathrm du$$
                                  Which simplifies to
                                  $$I=8intfrac{sqrt{sec^2u-1}}{sec^3u}tan u,mathrm du$$
                                  Noting that $sec^2u-1=tan^2u$, we can simplify more:
                                  $$I=8intfrac{tan^2u}{sec^3u}mathrm du$$
                                  Using the reciprocal and quotient identities this simplifies to
                                  $$I=8intsin^2ucos u,mathrm du$$
                                  Then we preform another substitution
                                  $$w=sin uRightarrow mathrm dw=cos u, mathrm du$$
                                  So we have
                                  $$I=8int w^2mathrm dw$$
                                  $$I=frac83w^3$$
                                  $$I=frac83sin^3u$$
                                  $$I=frac83(1-cos^2u)^{3/2}$$
                                  $$I=frac83bigg(1-frac1{sec^2u}bigg)^{3/2}$$
                                  since $x=frac12sec u$,
                                  $$I=frac83bigg(1-frac1{4x^2}bigg)^{3/2}$$
                                  $$I=frac{(4x^2-1)^{3/2}}{3x^3}+C$$






                                  share|cite|improve this answer
























                                    2












                                    2








                                    2






                                    This one is actually pretty neat
                                    $$I=intfrac{sqrt{4x^2-1}}{x^4}mathrm dx$$
                                    The first thing to do is get rid of the square root, so we do
                                    $$x=frac12sec uRightarrow mathrm dx=frac12sec utan u,mathrm du$$
                                    Hence
                                    $$I=frac12intfrac{sqrt{4cdotfrac14sec^2u-1}}{frac1{2^4}sec^4u}sec utan u,mathrm du$$
                                    Which simplifies to
                                    $$I=8intfrac{sqrt{sec^2u-1}}{sec^3u}tan u,mathrm du$$
                                    Noting that $sec^2u-1=tan^2u$, we can simplify more:
                                    $$I=8intfrac{tan^2u}{sec^3u}mathrm du$$
                                    Using the reciprocal and quotient identities this simplifies to
                                    $$I=8intsin^2ucos u,mathrm du$$
                                    Then we preform another substitution
                                    $$w=sin uRightarrow mathrm dw=cos u, mathrm du$$
                                    So we have
                                    $$I=8int w^2mathrm dw$$
                                    $$I=frac83w^3$$
                                    $$I=frac83sin^3u$$
                                    $$I=frac83(1-cos^2u)^{3/2}$$
                                    $$I=frac83bigg(1-frac1{sec^2u}bigg)^{3/2}$$
                                    since $x=frac12sec u$,
                                    $$I=frac83bigg(1-frac1{4x^2}bigg)^{3/2}$$
                                    $$I=frac{(4x^2-1)^{3/2}}{3x^3}+C$$






                                    share|cite|improve this answer












                                    This one is actually pretty neat
                                    $$I=intfrac{sqrt{4x^2-1}}{x^4}mathrm dx$$
                                    The first thing to do is get rid of the square root, so we do
                                    $$x=frac12sec uRightarrow mathrm dx=frac12sec utan u,mathrm du$$
                                    Hence
                                    $$I=frac12intfrac{sqrt{4cdotfrac14sec^2u-1}}{frac1{2^4}sec^4u}sec utan u,mathrm du$$
                                    Which simplifies to
                                    $$I=8intfrac{sqrt{sec^2u-1}}{sec^3u}tan u,mathrm du$$
                                    Noting that $sec^2u-1=tan^2u$, we can simplify more:
                                    $$I=8intfrac{tan^2u}{sec^3u}mathrm du$$
                                    Using the reciprocal and quotient identities this simplifies to
                                    $$I=8intsin^2ucos u,mathrm du$$
                                    Then we preform another substitution
                                    $$w=sin uRightarrow mathrm dw=cos u, mathrm du$$
                                    So we have
                                    $$I=8int w^2mathrm dw$$
                                    $$I=frac83w^3$$
                                    $$I=frac83sin^3u$$
                                    $$I=frac83(1-cos^2u)^{3/2}$$
                                    $$I=frac83bigg(1-frac1{sec^2u}bigg)^{3/2}$$
                                    since $x=frac12sec u$,
                                    $$I=frac83bigg(1-frac1{4x^2}bigg)^{3/2}$$
                                    $$I=frac{(4x^2-1)^{3/2}}{3x^3}+C$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 2 days ago









                                    clathratus

                                    3,017330




                                    3,017330















                                        Popular posts from this blog

                                        Human spaceflight

                                        Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                        張江高科駅