Trig Subsitution When There's No Square RootIntegral of $int sqrt1-4x^2$Struggling with an integral with trig substitutionUsing trig substitution, how do you solve an integral when the leading coefficient under the radical isn't 1?How do you solve for bounds when performing trig substitution and knowing solving the trig functions yields multiple correct values of $theta$?Domain of a square root natural log functionHelp with an integral of binomial differentialTrig substitution for $int fracx^2dxsqrt4 - x^2$Trigonometric integral with square root (residue theorem)Trig Subs issuesNot getting the right answer with alternate completing the square method on $intfracx^2sqrt3+4x-4x^2^3dx$

From an axiomatic set theoric approach why can we take uncountable unions?

When a wind turbine does not produce enough electricity how does the power company compensate for the loss?

Is it a Cyclops number? "Nobody" knows!

Shifting between bemols and diesis in the key signature

Vocabulary for giving just numbers, not a full answer

Is it possible that a question has only two answers?

Source permutation

Making a kiddush for a girl that has hard time finding shidduch

MySQL importing CSV files really slow

Street obstacles in New Zealand

Why restrict private health insurance?

Are all players supposed to be able to see each others' character sheets?

What will happen if my luggage gets delayed?

Does Christianity allow for believing on someone else's behalf?

How do we create new idioms and use them in a novel?

How to write a chaotic neutral protagonist and prevent my readers from thinking they are evil?

Do cubics always have one real root?

Is it safe to abruptly remove Arduino power?

Finitely many repeated replacements

Why is a very small peak with larger m/z not considered to be the molecular ion?

What do *foreign films* mean for an American?

Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?

Doubts in understanding some concepts of potential energy

PTIJ: Why does only a Shor Tam ask at the Seder, and not a Shor Mu'ad?



Trig Subsitution When There's No Square Root


Integral of $int sqrt1-4x^2$Struggling with an integral with trig substitutionUsing trig substitution, how do you solve an integral when the leading coefficient under the radical isn't 1?How do you solve for bounds when performing trig substitution and knowing solving the trig functions yields multiple correct values of $theta$?Domain of a square root natural log functionHelp with an integral of binomial differentialTrig substitution for $int fracx^2dxsqrt4 - x^2$Trigonometric integral with square root (residue theorem)Trig Subs issuesNot getting the right answer with alternate completing the square method on $intfracx^2sqrt3+4x-4x^2^3dx$













2












$begingroup$


I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.



I'm currently trying to solve the following question:



$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$



Anyway, so far, I have that:



$x = rtan theta$



$dx = rsec^2 theta$



$sqrt (r^2+x^2) = rsectheta$



Please click here to see the triangle I based the above values on.



.



Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.



= $Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta$



= $Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta$



= $fracAr int_a^b frac1sec^4thetadtheta$



= $fracAr int_a^b cos^4theta dtheta$



= $fracAr int_a^b (cos^2theta)^2 dtheta$



= $fracAr int_a^b [ frac121+cos(2theta)) ]^2dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$ which is wrong after I make substitutions.



.



I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.



It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^3/2quad$, life would be awesome.



I already know that the final answer is $fracAr(1-fracasqrtr^2+a^2)$, but I really want to understand this. An explanation would be most welcome!



Thank you in advance for all your help!










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
    $endgroup$
    – Kay K.
    4 hours ago
















2












$begingroup$


I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.



I'm currently trying to solve the following question:



$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$



Anyway, so far, I have that:



$x = rtan theta$



$dx = rsec^2 theta$



$sqrt (r^2+x^2) = rsectheta$



Please click here to see the triangle I based the above values on.



.



Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.



= $Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta$



= $Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta$



= $fracAr int_a^b frac1sec^4thetadtheta$



= $fracAr int_a^b cos^4theta dtheta$



= $fracAr int_a^b (cos^2theta)^2 dtheta$



= $fracAr int_a^b [ frac121+cos(2theta)) ]^2dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$ which is wrong after I make substitutions.



.



I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.



It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^3/2quad$, life would be awesome.



I already know that the final answer is $fracAr(1-fracasqrtr^2+a^2)$, but I really want to understand this. An explanation would be most welcome!



Thank you in advance for all your help!










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
    $endgroup$
    – Kay K.
    4 hours ago














2












2








2





$begingroup$


I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.



I'm currently trying to solve the following question:



$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$



Anyway, so far, I have that:



$x = rtan theta$



$dx = rsec^2 theta$



$sqrt (r^2+x^2) = rsectheta$



Please click here to see the triangle I based the above values on.



.



Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.



= $Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta$



= $Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta$



= $fracAr int_a^b frac1sec^4thetadtheta$



= $fracAr int_a^b cos^4theta dtheta$



= $fracAr int_a^b (cos^2theta)^2 dtheta$



= $fracAr int_a^b [ frac121+cos(2theta)) ]^2dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$ which is wrong after I make substitutions.



.



I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.



It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^3/2quad$, life would be awesome.



I already know that the final answer is $fracAr(1-fracasqrtr^2+a^2)$, but I really want to understand this. An explanation would be most welcome!



Thank you in advance for all your help!










share|cite|improve this question









$endgroup$




I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.



I'm currently trying to solve the following question:



$Ar int_a^infty fracdx(r^2+x^2)^(3/2)$



Anyway, so far, I have that:



$x = rtan theta$



$dx = rsec^2 theta$



$sqrt (r^2+x^2) = rsectheta$



Please click here to see the triangle I based the above values on.



.



Given that $(r^2+x^2)^(3/2)$ can be rewritten as $ (sqrtr^2+x^2)^3$, I begin to solve.
Please pretend I have $lim limits_b to infty$ in front of every line please.



= $Ar int_a^b fracrsec^2theta(rsectheta)^3dtheta$



= $Ar int_a^b fracrsec^2thetar^3sec^6thetadtheta$



= $fracAr int_a^b frac1sec^4thetadtheta$



= $fracAr int_a^b cos^4theta dtheta$



= $fracAr int_a^b (cos^2theta)^2 dtheta$



= $fracAr int_a^b [ frac121+cos(2theta)) ]^2dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $fracA4r int_a^b 1 + 2cos(2theta) dtheta quad+quad fracA4r int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $fracA4r[2theta+sin(2theta)] + fracA32r [4theta+sin(4theta)]$ which is wrong after I make substitutions.



.



I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.



It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^3/2quad$, life would be awesome.



I already know that the final answer is $fracAr(1-fracasqrtr^2+a^2)$, but I really want to understand this. An explanation would be most welcome!



Thank you in advance for all your help!







calculus integration improper-integrals trigonometric-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









CodingMeeCodingMee

204




204







  • 2




    $begingroup$
    The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
    $endgroup$
    – Kay K.
    4 hours ago













  • 2




    $begingroup$
    The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
    $endgroup$
    – Kay K.
    4 hours ago








2




2




$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago





$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago











2 Answers
2






active

oldest

votes


















4












$begingroup$

You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)




There's a slicker way to do it.



Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
$$

Now let's concentrate on the antiderivative
$$
intfrac1(1+u^2)^3/2,du=
intfrac1+u^2-u^2(1+u^2)^3/2,du=
intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
$$

Do the second term by parts
$$
int ufracu(1+u^2)^3/2,du=
-fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
$$

See what happens?
$$
intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
$$

which we can verify by direct differentiation.



Now
$$
left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
=1-fraca(r^2+a^2)^1/2
$$

and your integral is indeed
$$
fracArleft(1-fracasqrtr^2+a^2right)
$$






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    Firstly you made an error in the first line of working
    $$(rsec(theta))^3=r^3sec^3(theta)$$
    Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.






    share|cite|improve this answer









    $endgroup$












      Your Answer





      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142908%2ftrig-subsitution-when-theres-no-square-root%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)




      There's a slicker way to do it.



      Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
      $$
      fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
      $$

      Now let's concentrate on the antiderivative
      $$
      intfrac1(1+u^2)^3/2,du=
      intfrac1+u^2-u^2(1+u^2)^3/2,du=
      intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
      $$

      Do the second term by parts
      $$
      int ufracu(1+u^2)^3/2,du=
      -fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
      $$

      See what happens?
      $$
      intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
      $$

      which we can verify by direct differentiation.



      Now
      $$
      left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
      =1-fraca(r^2+a^2)^1/2
      $$

      and your integral is indeed
      $$
      fracArleft(1-fracasqrtr^2+a^2right)
      $$






      share|cite|improve this answer









      $endgroup$

















        4












        $begingroup$

        You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)




        There's a slicker way to do it.



        Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
        $$
        fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
        $$

        Now let's concentrate on the antiderivative
        $$
        intfrac1(1+u^2)^3/2,du=
        intfrac1+u^2-u^2(1+u^2)^3/2,du=
        intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
        $$

        Do the second term by parts
        $$
        int ufracu(1+u^2)^3/2,du=
        -fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
        $$

        See what happens?
        $$
        intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
        $$

        which we can verify by direct differentiation.



        Now
        $$
        left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
        =1-fraca(r^2+a^2)^1/2
        $$

        and your integral is indeed
        $$
        fracArleft(1-fracasqrtr^2+a^2right)
        $$






        share|cite|improve this answer









        $endgroup$















          4












          4








          4





          $begingroup$

          You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)




          There's a slicker way to do it.



          Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
          $$
          fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
          $$

          Now let's concentrate on the antiderivative
          $$
          intfrac1(1+u^2)^3/2,du=
          intfrac1+u^2-u^2(1+u^2)^3/2,du=
          intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
          $$

          Do the second term by parts
          $$
          int ufracu(1+u^2)^3/2,du=
          -fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
          $$

          See what happens?
          $$
          intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
          $$

          which we can verify by direct differentiation.



          Now
          $$
          left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
          =1-fraca(r^2+a^2)^1/2
          $$

          and your integral is indeed
          $$
          fracArleft(1-fracasqrtr^2+a^2right)
          $$






          share|cite|improve this answer









          $endgroup$



          You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)




          There's a slicker way to do it.



          Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
          $$
          fracArint_a/r^inftyfrac1(1+u^2)^3/2,du
          $$

          Now let's concentrate on the antiderivative
          $$
          intfrac1(1+u^2)^3/2,du=
          intfrac1+u^2-u^2(1+u^2)^3/2,du=
          intfrac1(1+u^2)^1/2,du-intfracu^2(1+u^2)^3/2,du
          $$

          Do the second term by parts
          $$
          int ufracu(1+u^2)^3/2,du=
          -fracu(1+u^2)^1/2+intfrac1(1+u^2)^1/2,du
          $$

          See what happens?
          $$
          intfrac1(1+u^2)^3/2,du=fracu(1+u^2)^1/2+c
          $$

          which we can verify by direct differentiation.



          Now
          $$
          left[fracu(1+u^2)^1/2right]_a/r^infty=1-fraca/r(1+(a/r)^2)^1/2
          =1-fraca(r^2+a^2)^1/2
          $$

          and your integral is indeed
          $$
          fracArleft(1-fracasqrtr^2+a^2right)
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          egregegreg

          184k1486205




          184k1486205





















              3












              $begingroup$

              Firstly you made an error in the first line of working
              $$(rsec(theta))^3=r^3sec^3(theta)$$
              Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.






              share|cite|improve this answer









              $endgroup$

















                3












                $begingroup$

                Firstly you made an error in the first line of working
                $$(rsec(theta))^3=r^3sec^3(theta)$$
                Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.






                share|cite|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  Firstly you made an error in the first line of working
                  $$(rsec(theta))^3=r^3sec^3(theta)$$
                  Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.






                  share|cite|improve this answer









                  $endgroup$



                  Firstly you made an error in the first line of working
                  $$(rsec(theta))^3=r^3sec^3(theta)$$
                  Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan(fracxr)$ then the limits should change as $x=a implies theta=arctan(fracar)$ also $x=infty implies theta=fracpi2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Peter ForemanPeter Foreman

                  3,4421216




                  3,4421216



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142908%2ftrig-subsitution-when-theres-no-square-root%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Möglingen Índice Localización Historia Demografía Referencias Enlaces externos Menú de navegación48°53′18″N 9°07′45″E / 48.888333333333, 9.129166666666748°53′18″N 9°07′45″E / 48.888333333333, 9.1291666666667Sitio web oficial Mapa de Möglingen«Gemeinden in Deutschland nach Fläche, Bevölkerung und Postleitzahl am 30.09.2016»Möglingen

                      Virtualbox - Configuration error: Querying “UUID” failed (VERR_CFGM_VALUE_NOT_FOUND)“VERR_SUPLIB_WORLD_WRITABLE” error when trying to installing OS in virtualboxVirtual Box Kernel errorFailed to open a seesion for the virtual machineFailed to open a session for the virtual machineUbuntu 14.04 LTS Virtualbox errorcan't use VM VirtualBoxusing virtualboxI can't run Linux-64 Bit on VirtualBoxUnable to insert the virtual optical disk (VBoxguestaddition) in virtual machine for ubuntu server in win 10VirtuaBox in Ubuntu 18.04 Issues with Win10.ISO Installation

                      Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org