Integral of $frac{x^2}{sqrt{x^2+5}}$












1












$begingroup$


I need help with this integral:



$$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






calculus integration trigonometry







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$endgroup$

















    1












    $begingroup$


    I need help with this integral:



    $$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



    I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
    For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






    calculus integration trigonometry







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I need help with this integral:



      $$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



      I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
      For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






      calculus integration trigonometry







      share|cite|improve this question











      $endgroup$




      I need help with this integral:



      $$I = int frac{x^2}{sqrt{x^2+5}}, dx$$



      I substituted $x = sqrt{5}tan{theta}$, and reached $$I = 5int frac{sin{(theta)}^2}{cos{(theta)}^3},dtheta$$ which I'm unable to solve.
      For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.






      calculus integration trigonometry




      calculus integration trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 13:28









      Robert Z

      102k1072145




      102k1072145










      asked Jan 18 at 12:55









      Yizhar AmirYizhar Amir

      17217




      17217






















          6 Answers
          6






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          3












          $begingroup$

          Hint: (Use hyperbolic trigonometric substitution to)
          $$
          frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
          $$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
              $endgroup$
              – GEdgar
              Jan 18 at 13:08



















            2












            $begingroup$

            Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
            $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
            Finally note that $t=x+sqrt{x^2+5}$.






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                share|cite|improve this answer









                $endgroup$





















                  0












                  $begingroup$

                  What about



                  $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                  $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                  $$intleft(5sinh^2u-10sinh u+1right)du$$



                  and now it is almost an immediate integral.






                  share|cite|improve this answer









                  $endgroup$














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                    6 Answers
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                    active

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                    6 Answers
                    6






                    active

                    oldest

                    votes









                    active

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                    votes






                    active

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                    3












                    $begingroup$

                    Hint: (Use hyperbolic trigonometric substitution to)
                    $$
                    frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                    $$






                    share|cite|improve this answer









                    $endgroup$


















                      3












                      $begingroup$

                      Hint: (Use hyperbolic trigonometric substitution to)
                      $$
                      frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                      $$






                      share|cite|improve this answer









                      $endgroup$
















                        3












                        3








                        3





                        $begingroup$

                        Hint: (Use hyperbolic trigonometric substitution to)
                        $$
                        frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                        $$






                        share|cite|improve this answer









                        $endgroup$



                        Hint: (Use hyperbolic trigonometric substitution to)
                        $$
                        frac{x^2}{sqrt{x^2+5}}=sqrt{x^2+5}-frac{5}{sqrt{x^2+5}}.
                        $$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 18 at 12:59









                        SongSong

                        18.6k21651




                        18.6k21651























                            2












                            $begingroup$

                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






                            share|cite|improve this answer









                            $endgroup$













                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08
















                            2












                            $begingroup$

                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






                            share|cite|improve this answer









                            $endgroup$













                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08














                            2












                            2








                            2





                            $begingroup$

                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.






                            share|cite|improve this answer









                            $endgroup$



                            Note that$$frac{sin^2(theta)}{cos^3(theta)}=frac{sin^2(theta)cos(theta)}{bigl(1-sin^2(theta)bigr)^2}.$$So, you can do $sin(theta)=u$ and $cos(theta),mathrm dtheta=mathrm du$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 18 at 13:03









                            José Carlos SantosJosé Carlos Santos

                            176k24135244




                            176k24135244












                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08


















                            • $begingroup$
                              This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                              $endgroup$
                              – GEdgar
                              Jan 18 at 13:08
















                            $begingroup$
                            This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                            $endgroup$
                            – GEdgar
                            Jan 18 at 13:08




                            $begingroup$
                            This method works when you have power of $sin$ times power of $cos$ where at least one of the exponents is odd.
                            $endgroup$
                            – GEdgar
                            Jan 18 at 13:08











                            2












                            $begingroup$

                            Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                            $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                            Finally note that $t=x+sqrt{x^2+5}$.






                            share|cite|improve this answer











                            $endgroup$


















                              2












                              $begingroup$

                              Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                              $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                              Finally note that $t=x+sqrt{x^2+5}$.






                              share|cite|improve this answer











                              $endgroup$
















                                2












                                2








                                2





                                $begingroup$

                                Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                                $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                                Finally note that $t=x+sqrt{x^2+5}$.






                                share|cite|improve this answer











                                $endgroup$



                                Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate):
                                $$int frac{x^2}{sqrt{x^2+5}}, dx=int left(frac{25}{4t^3}+frac{t}{4}-frac{5}{2t}right), dt.$$
                                Finally note that $t=x+sqrt{x^2+5}$.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jan 18 at 13:26

























                                answered Jan 18 at 13:08









                                Robert ZRobert Z

                                102k1072145




                                102k1072145























                                    1












                                    $begingroup$

                                    Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      1












                                      $begingroup$

                                      Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Hint: Rewrite $intfrac{sin^2{(theta)}}{cos^3{(theta)}} d theta$ as $int tan^2 theta sec theta d theta$. Let $u = sec theta$; then $du = ?$ - this will reduce the integral to something trivial.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jan 18 at 13:43









                                        bjcolby15bjcolby15

                                        1,51711016




                                        1,51711016























                                            0












                                            $begingroup$

                                            Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.






                                                share|cite|improve this answer









                                                $endgroup$



                                                Hint: Substitute $y=sqrt{x^2+5}$. Then $dy= frac{x}{sqrt{x^2+5}}dx$, so the integral is $int sqrt{y^2-5} dy$.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jan 18 at 13:01









                                                A. PongráczA. Pongrácz

                                                6,1171929




                                                6,1171929























                                                    0












                                                    $begingroup$

                                                    What about



                                                    $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                    $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                    $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                    and now it is almost an immediate integral.






                                                    share|cite|improve this answer









                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      What about



                                                      $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                      $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                      $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                      and now it is almost an immediate integral.






                                                      share|cite|improve this answer









                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        What about



                                                        $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                        $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                        $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                        and now it is almost an immediate integral.






                                                        share|cite|improve this answer









                                                        $endgroup$



                                                        What about



                                                        $$1+frac x{sqrt5}=sinh uimplies dx=cosh usqrt 5,duimplies$$



                                                        $$intfrac{x^2}{sqrt{x^2+5}}==frac1{sqrt5}intfrac{x^2}{sqrt{1+left(frac x{sqrt 5}right)^2}}dx=frac1{sqrt5}intfrac{(sqrt5,sinh u-1)^2}{cosh u}sqrt5,cosh u,du=$$



                                                        $$intleft(5sinh^2u-10sinh u+1right)du$$



                                                        and now it is almost an immediate integral.







                                                        share|cite|improve this answer












                                                        share|cite|improve this answer



                                                        share|cite|improve this answer










                                                        answered Jan 18 at 13:03









                                                        DonAntonioDonAntonio

                                                        180k1495233




                                                        180k1495233






























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