Trig Subsitution When There's No Square RootIntegral of $int sqrt{1-4x^2}$Struggling with an integral with...

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Trig Subsitution When There's No Square Root

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Trig Subsitution When There's No Square Root


Integral of $int sqrt{1-4x^2}$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 $int frac{x^2dx}{sqrt{4 - x^2}}$Trigonometric integral with square root (residue theorem)Trig Subs issuesNot getting the right answer with alternate completing the square method on $intfrac{x^2}{sqrt{3+4x-4x^2}^3}dx$













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:



$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$



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 $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.



= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$



= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$



= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$



= $frac{A}{r} int_a^b cos^4theta dtheta$



= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$



= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ 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 $quad ^{3/2}quad$, life would be awesome.



I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, 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









$endgroup$








  • 2




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



$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$



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 $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.



= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$



= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$



= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$



= $frac{A}{r} int_a^b cos^4theta dtheta$



= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$



= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ 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 $quad ^{3/2}quad$, life would be awesome.



I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, 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









$endgroup$








  • 2




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



$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$



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 $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.



= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$



= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$



= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$



= $frac{A}{r} int_a^b cos^4theta dtheta$



= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$



= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ 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 $quad ^{3/2}quad$, life would be awesome.



I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, 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









$endgroup$




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:



$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$



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 $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.



= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$



= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$



= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$



= $frac{A}{r} int_a^b cos^4theta dtheta$



= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$



= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$



= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$



And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ 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 $quad ^{3/2}quad$, life would be awesome.



I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, 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




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 3 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.
    2 hours ago
















  • 2




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










2




2




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






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












2 Answers
2






active

oldest

votes


















3












$begingroup$

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
$$
frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
$$
Now let's concentrate on the antiderivative
$$
intfrac{1}{(1+u^2)^{3/2}},du=
intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
$$
Do the second term by parts
$$
int ufrac{u}{(1+u^2)^{3/2}},du=
-frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
$$
See what happens?
$$
intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
$$
which we can verify by direct differentiation.



Now
$$
left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
=1-frac{a}{(r^2+a^2)^{1/2}}
$$
and your integral is indeed
$$
frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
$$






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 $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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
      $$
      frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
      $$
      Now let's concentrate on the antiderivative
      $$
      intfrac{1}{(1+u^2)^{3/2}},du=
      intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
      intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
      $$
      Do the second term by parts
      $$
      int ufrac{u}{(1+u^2)^{3/2}},du=
      -frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
      $$
      See what happens?
      $$
      intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
      $$
      which we can verify by direct differentiation.



      Now
      $$
      left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
      =1-frac{a}{(r^2+a^2)^{1/2}}
      $$
      and your integral is indeed
      $$
      frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
      $$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        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
        $$
        frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
        $$
        Now let's concentrate on the antiderivative
        $$
        intfrac{1}{(1+u^2)^{3/2}},du=
        intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
        intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
        $$
        Do the second term by parts
        $$
        int ufrac{u}{(1+u^2)^{3/2}},du=
        -frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
        $$
        See what happens?
        $$
        intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
        $$
        which we can verify by direct differentiation.



        Now
        $$
        left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
        =1-frac{a}{(r^2+a^2)^{1/2}}
        $$
        and your integral is indeed
        $$
        frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
        $$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          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
          $$
          frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
          $$
          Now let's concentrate on the antiderivative
          $$
          intfrac{1}{(1+u^2)^{3/2}},du=
          intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
          intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
          $$
          Do the second term by parts
          $$
          int ufrac{u}{(1+u^2)^{3/2}},du=
          -frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
          $$
          See what happens?
          $$
          intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
          $$
          which we can verify by direct differentiation.



          Now
          $$
          left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
          =1-frac{a}{(r^2+a^2)^{1/2}}
          $$
          and your integral is indeed
          $$
          frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
          $$






          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
          $$
          frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
          $$
          Now let's concentrate on the antiderivative
          $$
          intfrac{1}{(1+u^2)^{3/2}},du=
          intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
          intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
          $$
          Do the second term by parts
          $$
          int ufrac{u}{(1+u^2)^{3/2}},du=
          -frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
          $$
          See what happens?
          $$
          intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
          $$
          which we can verify by direct differentiation.



          Now
          $$
          left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
          =1-frac{a}{(r^2+a^2)^{1/2}}
          $$
          and your integral is indeed
          $$
          frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 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 $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.






              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 $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.






                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 $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.






                  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 $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Peter ForemanPeter Foreman

                  3,4421216




                  3,4421216






























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