Primitive of a function with $sin frac{1}{x}$












2












$begingroup$


I have the next integral:
$$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
Can somebody give me some tips about what should I do next, please?






real-analysis calculus trigonometry indefinite-integrals substitution







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    2












    $begingroup$


    I have the next integral:
    $$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
    I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
    Can somebody give me some tips about what should I do next, please?






    real-analysis calculus trigonometry indefinite-integrals substitution







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I have the next integral:
      $$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
      I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
      Can somebody give me some tips about what should I do next, please?






      real-analysis calculus trigonometry indefinite-integrals substitution







      share|cite|improve this question











      $endgroup$




      I have the next integral:
      $$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
      I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
      Can somebody give me some tips about what should I do next, please?






      real-analysis calculus trigonometry indefinite-integrals substitution




      real-analysis calculus trigonometry indefinite-integrals substitution






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      share|cite|improve this question













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      edited Jan 16 at 21:24









      Bernard

      121k740116




      121k740116










      asked Jan 16 at 20:19









      GaboruGaboru

      1047




      1047






















          2 Answers
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          $begingroup$

          The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
          $$
          intfrac{-sin u}{sqrt{4+3sin2u}},du=
          intfrac{-sin u}{sqrt{4+3sin2u}},du
          $$
          This can be improved by setting $u=pi/4-v$, so we get
          $$
          frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
          frac{1}{sqrt{2}}biggl(
          intfrac{cos v}{sqrt{7-6sin^2v}},dv
          -intfrac{sin v}{sqrt{6cos^2v+1}},dv
          biggr)
          $$
          that you should be able to manage.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            As $(sin vpmcos v)^2=1pmsin2v$



            $$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$



            As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for



            $$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$



            Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$






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              2 Answers
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              2 Answers
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              active

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              1












              $begingroup$

              The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
              $$
              intfrac{-sin u}{sqrt{4+3sin2u}},du=
              intfrac{-sin u}{sqrt{4+3sin2u}},du
              $$
              This can be improved by setting $u=pi/4-v$, so we get
              $$
              frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
              frac{1}{sqrt{2}}biggl(
              intfrac{cos v}{sqrt{7-6sin^2v}},dv
              -intfrac{sin v}{sqrt{6cos^2v+1}},dv
              biggr)
              $$
              that you should be able to manage.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
                $$
                intfrac{-sin u}{sqrt{4+3sin2u}},du=
                intfrac{-sin u}{sqrt{4+3sin2u}},du
                $$
                This can be improved by setting $u=pi/4-v$, so we get
                $$
                frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
                frac{1}{sqrt{2}}biggl(
                intfrac{cos v}{sqrt{7-6sin^2v}},dv
                -intfrac{sin v}{sqrt{6cos^2v+1}},dv
                biggr)
                $$
                that you should be able to manage.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
                  $$
                  intfrac{-sin u}{sqrt{4+3sin2u}},du=
                  intfrac{-sin u}{sqrt{4+3sin2u}},du
                  $$
                  This can be improved by setting $u=pi/4-v$, so we get
                  $$
                  frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
                  frac{1}{sqrt{2}}biggl(
                  intfrac{cos v}{sqrt{7-6sin^2v}},dv
                  -intfrac{sin v}{sqrt{6cos^2v+1}},dv
                  biggr)
                  $$
                  that you should be able to manage.






                  share|cite|improve this answer









                  $endgroup$



                  The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
                  $$
                  intfrac{-sin u}{sqrt{4+3sin2u}},du=
                  intfrac{-sin u}{sqrt{4+3sin2u}},du
                  $$
                  This can be improved by setting $u=pi/4-v$, so we get
                  $$
                  frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
                  frac{1}{sqrt{2}}biggl(
                  intfrac{cos v}{sqrt{7-6sin^2v}},dv
                  -intfrac{sin v}{sqrt{6cos^2v+1}},dv
                  biggr)
                  $$
                  that you should be able to manage.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 16 at 21:14









                  egregegreg

                  182k1486204




                  182k1486204























                      0












                      $begingroup$

                      As $(sin vpmcos v)^2=1pmsin2v$



                      $$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$



                      As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for



                      $$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$



                      Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        As $(sin vpmcos v)^2=1pmsin2v$



                        $$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$



                        As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for



                        $$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$



                        Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          As $(sin vpmcos v)^2=1pmsin2v$



                          $$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$



                          As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for



                          $$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$



                          Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$






                          share|cite|improve this answer









                          $endgroup$



                          As $(sin vpmcos v)^2=1pmsin2v$



                          $$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$



                          As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for



                          $$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$



                          Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 17 at 7:55









                          lab bhattacharjeelab bhattacharjee

                          226k15157275




                          226k15157275






























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