Don't understand how to use trig sub on $intfrac{x^3}{(4x^2+9)^frac{3}{2}}dx$












2












$begingroup$


In my textbook it says




First we note that $(4x^2+9)^frac{3}{2} = (sqrt{4x^2+9})^3$ so trigonometric substitution is appropriate. Although $sqrt{4x^2+9}$ is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution $u=2x$. When we combine this with the tangent substitution, we have $x=frac{3}{2}tan(theta)$, which gives $dx=frac{3}{2}sec^2theta dtheta$




I understand the basic trig substitutions in integrals, and I recognize that this is going to turn into some kind of $tan$ substitution since the denominator $sqrt{4x^2+9}$ is equivalent to $sqrt{x^2+a^2}$, but for the life of me I cannot figure out why he is subbing $u=2x$. Does it have something to do with the fact that $sqrt{4x^2}=2x$? After that, then he subs in $x=frac{3}{2}tan(theta)$ which I don't see relating to anything except for that maybe when the $4$ gets factored out of the radical the $9$ somehow becomes $9/2$? The only relation to $frac{3}{2}$ in this integral I see is the denominator's exponent, which I doubt is actually the reason for subbing $x=frac{3}{2}tan(theta)$



I'm not asking for the entire problem to be solved, I know I can do that looking at my textbook as the parts after this "preliminary" step make sense, I just have no idea how to start this. If I'm missing some obvious algebraic step I'm sorry, but if anyone can shed some light on what's happening with the $u=2x$ and the $tan$ sub, I would appreciate it so so much.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    theta gives $theta$
    $endgroup$
    – Shrey Joshi
    Jan 5 at 21:26






  • 1




    $begingroup$
    What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$?
    $endgroup$
    – Ennar
    Jan 5 at 22:00
















2












$begingroup$


In my textbook it says




First we note that $(4x^2+9)^frac{3}{2} = (sqrt{4x^2+9})^3$ so trigonometric substitution is appropriate. Although $sqrt{4x^2+9}$ is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution $u=2x$. When we combine this with the tangent substitution, we have $x=frac{3}{2}tan(theta)$, which gives $dx=frac{3}{2}sec^2theta dtheta$




I understand the basic trig substitutions in integrals, and I recognize that this is going to turn into some kind of $tan$ substitution since the denominator $sqrt{4x^2+9}$ is equivalent to $sqrt{x^2+a^2}$, but for the life of me I cannot figure out why he is subbing $u=2x$. Does it have something to do with the fact that $sqrt{4x^2}=2x$? After that, then he subs in $x=frac{3}{2}tan(theta)$ which I don't see relating to anything except for that maybe when the $4$ gets factored out of the radical the $9$ somehow becomes $9/2$? The only relation to $frac{3}{2}$ in this integral I see is the denominator's exponent, which I doubt is actually the reason for subbing $x=frac{3}{2}tan(theta)$



I'm not asking for the entire problem to be solved, I know I can do that looking at my textbook as the parts after this "preliminary" step make sense, I just have no idea how to start this. If I'm missing some obvious algebraic step I'm sorry, but if anyone can shed some light on what's happening with the $u=2x$ and the $tan$ sub, I would appreciate it so so much.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    theta gives $theta$
    $endgroup$
    – Shrey Joshi
    Jan 5 at 21:26






  • 1




    $begingroup$
    What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$?
    $endgroup$
    – Ennar
    Jan 5 at 22:00














2












2








2





$begingroup$


In my textbook it says




First we note that $(4x^2+9)^frac{3}{2} = (sqrt{4x^2+9})^3$ so trigonometric substitution is appropriate. Although $sqrt{4x^2+9}$ is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution $u=2x$. When we combine this with the tangent substitution, we have $x=frac{3}{2}tan(theta)$, which gives $dx=frac{3}{2}sec^2theta dtheta$




I understand the basic trig substitutions in integrals, and I recognize that this is going to turn into some kind of $tan$ substitution since the denominator $sqrt{4x^2+9}$ is equivalent to $sqrt{x^2+a^2}$, but for the life of me I cannot figure out why he is subbing $u=2x$. Does it have something to do with the fact that $sqrt{4x^2}=2x$? After that, then he subs in $x=frac{3}{2}tan(theta)$ which I don't see relating to anything except for that maybe when the $4$ gets factored out of the radical the $9$ somehow becomes $9/2$? The only relation to $frac{3}{2}$ in this integral I see is the denominator's exponent, which I doubt is actually the reason for subbing $x=frac{3}{2}tan(theta)$



I'm not asking for the entire problem to be solved, I know I can do that looking at my textbook as the parts after this "preliminary" step make sense, I just have no idea how to start this. If I'm missing some obvious algebraic step I'm sorry, but if anyone can shed some light on what's happening with the $u=2x$ and the $tan$ sub, I would appreciate it so so much.










share|cite|improve this question











$endgroup$




In my textbook it says




First we note that $(4x^2+9)^frac{3}{2} = (sqrt{4x^2+9})^3$ so trigonometric substitution is appropriate. Although $sqrt{4x^2+9}$ is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution $u=2x$. When we combine this with the tangent substitution, we have $x=frac{3}{2}tan(theta)$, which gives $dx=frac{3}{2}sec^2theta dtheta$




I understand the basic trig substitutions in integrals, and I recognize that this is going to turn into some kind of $tan$ substitution since the denominator $sqrt{4x^2+9}$ is equivalent to $sqrt{x^2+a^2}$, but for the life of me I cannot figure out why he is subbing $u=2x$. Does it have something to do with the fact that $sqrt{4x^2}=2x$? After that, then he subs in $x=frac{3}{2}tan(theta)$ which I don't see relating to anything except for that maybe when the $4$ gets factored out of the radical the $9$ somehow becomes $9/2$? The only relation to $frac{3}{2}$ in this integral I see is the denominator's exponent, which I doubt is actually the reason for subbing $x=frac{3}{2}tan(theta)$



I'm not asking for the entire problem to be solved, I know I can do that looking at my textbook as the parts after this "preliminary" step make sense, I just have no idea how to start this. If I'm missing some obvious algebraic step I'm sorry, but if anyone can shed some light on what's happening with the $u=2x$ and the $tan$ sub, I would appreciate it so so much.







calculus integration trigonometric-integrals






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edited Jan 5 at 23:13









M. Nestor

741113




741113










asked Jan 5 at 21:24









Ben DreslinskiBen Dreslinski

224




224








  • 1




    $begingroup$
    theta gives $theta$
    $endgroup$
    – Shrey Joshi
    Jan 5 at 21:26






  • 1




    $begingroup$
    What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$?
    $endgroup$
    – Ennar
    Jan 5 at 22:00














  • 1




    $begingroup$
    theta gives $theta$
    $endgroup$
    – Shrey Joshi
    Jan 5 at 21:26






  • 1




    $begingroup$
    What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$?
    $endgroup$
    – Ennar
    Jan 5 at 22:00








1




1




$begingroup$
theta gives $theta$
$endgroup$
– Shrey Joshi
Jan 5 at 21:26




$begingroup$
theta gives $theta$
$endgroup$
– Shrey Joshi
Jan 5 at 21:26




1




1




$begingroup$
What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$?
$endgroup$
– Ennar
Jan 5 at 22:00




$begingroup$
What substitution would you use for the case $u^2+9$? Now, by what substitution would you get $u^2+9$ from $4x^2+9$?
$endgroup$
– Ennar
Jan 5 at 22:00










3 Answers
3






active

oldest

votes


















3












$begingroup$

Probably the textbook first defines $u=2x$ and then $u=3tan w$ for some $w$ to avoid confusion (of course probably!), though here we merge the two substitutions together to obtain $$x={3over 2}tan theta$$which by substitution leads to $$int{x^3over (4x^2+9)sqrt{4x^2+9}}dx=int {{27over 8}tan^3thetaover 27(1+tan^2theta)sectheta}cdot {3over 2}(1+tan^2theta)dtheta\={3over 16}int{tan^3thetaover sec theta}dtheta$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So cases where the radicand's constants are not in the $sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks.
    $endgroup$
    – Ben Dreslinski
    Jan 5 at 23:35












  • $begingroup$
    You're welcome. That's right. For example a $sqrt{x^2-a^2}$ term requires a $x=acosh u$ substitution....
    $endgroup$
    – Mostafa Ayaz
    Jan 6 at 7:03



















3












$begingroup$

You could avoid the extra substitution altogether (and greatly simplify other trig substitution problems) if you construct a right triangle to inform your substitutions.



Imagine a right triangle whose opposite (to $theta$) leg has length $2x$, adjacent leg has length $3$, and hypotenuse has length $sqrt{4x^2 + 9}$. From this I can deduce that $x = frac{3}{2}tantheta$, and therefore $dx = frac{3}{2}sec^2theta dtheta$. I can also determine that $sqrt{4x^2 + 9} = 3sectheta$. Substituting all this in gives:
$$
intfrac{x^3}{(4x^2+9)^frac{3}{2}} = int frac{left(frac{3}{2}tanthetaright)^3}{left(3secthetaright)^3} cdot frac{3}{2}sec^2theta dtheta = frac{3}{16}intfrac{tan^3theta}{sectheta}dtheta.
$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    The answers provided so far already outline exactly how to approach your question with trigonometric substitutions. Here I will speak more broadly about the common trig substitutions. The general goal is to take some rational function of multiple terms and reduce it down to a single term. For instance, here you want to have $4x^2 + 9$ to be represented by a single term.



    This motivation of using trigonometry to reduce multiple terms to a single (or overall a much reduced amount) is quite common within integration. Although there are many different identities used, the majority are centred around the principle identity:



    begin{equation}
    sin^2(x) + cos^2(x) = 1
    end{equation}



    From here three core forms can be derived. The first is just a simple rearrangement:



    begin{equation}
    sin^2(x) + cos^2(x) = 1 rightarrow cos^2(x) = 1 - sin^2(x)
    end{equation}



    The next two are formed by dividing the principle identity through by $cos^2(x)$
    begin{equation}
    frac{1}{cos^2(x)}left(sin^2(x) + cos^2(x) right) = frac{1}{cos^2(x)} rightarrow tan^2(x) + 1 = sec^2(x)
    end{equation}



    Combining the three we have:




    1. $sec^2(x) = tan^2(x) + 1$

    2. $tan^2(x) = sec^2(x) - 1$

    3. $sin^2(x) = 1 - cos^2(x)$


    So here we two three different situations in which multiple terms can be reduced into a single term. These are (and to match the ordering of the list above)




    1. $a^2 + x^2$

    2. $x^2 - a^2$

    3. $a^2 - x^2$


    For your question you are working with (1).



    Edit - Sorry I should have said that not only are you looking to replace multiple terms with a single term but you are also often looking to get rid of a 'square root'. You will see in the list above that in all three cases the resultant term is a squared term and thus when taking the square root you get rid of it!






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Probably the textbook first defines $u=2x$ and then $u=3tan w$ for some $w$ to avoid confusion (of course probably!), though here we merge the two substitutions together to obtain $$x={3over 2}tan theta$$which by substitution leads to $$int{x^3over (4x^2+9)sqrt{4x^2+9}}dx=int {{27over 8}tan^3thetaover 27(1+tan^2theta)sectheta}cdot {3over 2}(1+tan^2theta)dtheta\={3over 16}int{tan^3thetaover sec theta}dtheta$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        So cases where the radicand's constants are not in the $sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks.
        $endgroup$
        – Ben Dreslinski
        Jan 5 at 23:35












      • $begingroup$
        You're welcome. That's right. For example a $sqrt{x^2-a^2}$ term requires a $x=acosh u$ substitution....
        $endgroup$
        – Mostafa Ayaz
        Jan 6 at 7:03
















      3












      $begingroup$

      Probably the textbook first defines $u=2x$ and then $u=3tan w$ for some $w$ to avoid confusion (of course probably!), though here we merge the two substitutions together to obtain $$x={3over 2}tan theta$$which by substitution leads to $$int{x^3over (4x^2+9)sqrt{4x^2+9}}dx=int {{27over 8}tan^3thetaover 27(1+tan^2theta)sectheta}cdot {3over 2}(1+tan^2theta)dtheta\={3over 16}int{tan^3thetaover sec theta}dtheta$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        So cases where the radicand's constants are not in the $sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks.
        $endgroup$
        – Ben Dreslinski
        Jan 5 at 23:35












      • $begingroup$
        You're welcome. That's right. For example a $sqrt{x^2-a^2}$ term requires a $x=acosh u$ substitution....
        $endgroup$
        – Mostafa Ayaz
        Jan 6 at 7:03














      3












      3








      3





      $begingroup$

      Probably the textbook first defines $u=2x$ and then $u=3tan w$ for some $w$ to avoid confusion (of course probably!), though here we merge the two substitutions together to obtain $$x={3over 2}tan theta$$which by substitution leads to $$int{x^3over (4x^2+9)sqrt{4x^2+9}}dx=int {{27over 8}tan^3thetaover 27(1+tan^2theta)sectheta}cdot {3over 2}(1+tan^2theta)dtheta\={3over 16}int{tan^3thetaover sec theta}dtheta$$






      share|cite|improve this answer









      $endgroup$



      Probably the textbook first defines $u=2x$ and then $u=3tan w$ for some $w$ to avoid confusion (of course probably!), though here we merge the two substitutions together to obtain $$x={3over 2}tan theta$$which by substitution leads to $$int{x^3over (4x^2+9)sqrt{4x^2+9}}dx=int {{27over 8}tan^3thetaover 27(1+tan^2theta)sectheta}cdot {3over 2}(1+tan^2theta)dtheta\={3over 16}int{tan^3thetaover sec theta}dtheta$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 5 at 21:43









      Mostafa AyazMostafa Ayaz

      15.3k3939




      15.3k3939












      • $begingroup$
        So cases where the radicand's constants are not in the $sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks.
        $endgroup$
        – Ben Dreslinski
        Jan 5 at 23:35












      • $begingroup$
        You're welcome. That's right. For example a $sqrt{x^2-a^2}$ term requires a $x=acosh u$ substitution....
        $endgroup$
        – Mostafa Ayaz
        Jan 6 at 7:03


















      • $begingroup$
        So cases where the radicand's constants are not in the $sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks.
        $endgroup$
        – Ben Dreslinski
        Jan 5 at 23:35












      • $begingroup$
        You're welcome. That's right. For example a $sqrt{x^2-a^2}$ term requires a $x=acosh u$ substitution....
        $endgroup$
        – Mostafa Ayaz
        Jan 6 at 7:03
















      $begingroup$
      So cases where the radicand's constants are not in the $sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks.
      $endgroup$
      – Ben Dreslinski
      Jan 5 at 23:35






      $begingroup$
      So cases where the radicand's constants are not in the $sqrt{x^2+a^2}$ (or any other trig sub format) tend to result in multiple $u$ substitutions leaving you to solve for $x$ by setting them equal? In that case, I suppose that's what the book meant when he said "combine," well now I feel silly; much thanks.
      $endgroup$
      – Ben Dreslinski
      Jan 5 at 23:35














      $begingroup$
      You're welcome. That's right. For example a $sqrt{x^2-a^2}$ term requires a $x=acosh u$ substitution....
      $endgroup$
      – Mostafa Ayaz
      Jan 6 at 7:03




      $begingroup$
      You're welcome. That's right. For example a $sqrt{x^2-a^2}$ term requires a $x=acosh u$ substitution....
      $endgroup$
      – Mostafa Ayaz
      Jan 6 at 7:03











      3












      $begingroup$

      You could avoid the extra substitution altogether (and greatly simplify other trig substitution problems) if you construct a right triangle to inform your substitutions.



      Imagine a right triangle whose opposite (to $theta$) leg has length $2x$, adjacent leg has length $3$, and hypotenuse has length $sqrt{4x^2 + 9}$. From this I can deduce that $x = frac{3}{2}tantheta$, and therefore $dx = frac{3}{2}sec^2theta dtheta$. I can also determine that $sqrt{4x^2 + 9} = 3sectheta$. Substituting all this in gives:
      $$
      intfrac{x^3}{(4x^2+9)^frac{3}{2}} = int frac{left(frac{3}{2}tanthetaright)^3}{left(3secthetaright)^3} cdot frac{3}{2}sec^2theta dtheta = frac{3}{16}intfrac{tan^3theta}{sectheta}dtheta.
      $$






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        You could avoid the extra substitution altogether (and greatly simplify other trig substitution problems) if you construct a right triangle to inform your substitutions.



        Imagine a right triangle whose opposite (to $theta$) leg has length $2x$, adjacent leg has length $3$, and hypotenuse has length $sqrt{4x^2 + 9}$. From this I can deduce that $x = frac{3}{2}tantheta$, and therefore $dx = frac{3}{2}sec^2theta dtheta$. I can also determine that $sqrt{4x^2 + 9} = 3sectheta$. Substituting all this in gives:
        $$
        intfrac{x^3}{(4x^2+9)^frac{3}{2}} = int frac{left(frac{3}{2}tanthetaright)^3}{left(3secthetaright)^3} cdot frac{3}{2}sec^2theta dtheta = frac{3}{16}intfrac{tan^3theta}{sectheta}dtheta.
        $$






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          You could avoid the extra substitution altogether (and greatly simplify other trig substitution problems) if you construct a right triangle to inform your substitutions.



          Imagine a right triangle whose opposite (to $theta$) leg has length $2x$, adjacent leg has length $3$, and hypotenuse has length $sqrt{4x^2 + 9}$. From this I can deduce that $x = frac{3}{2}tantheta$, and therefore $dx = frac{3}{2}sec^2theta dtheta$. I can also determine that $sqrt{4x^2 + 9} = 3sectheta$. Substituting all this in gives:
          $$
          intfrac{x^3}{(4x^2+9)^frac{3}{2}} = int frac{left(frac{3}{2}tanthetaright)^3}{left(3secthetaright)^3} cdot frac{3}{2}sec^2theta dtheta = frac{3}{16}intfrac{tan^3theta}{sectheta}dtheta.
          $$






          share|cite|improve this answer











          $endgroup$



          You could avoid the extra substitution altogether (and greatly simplify other trig substitution problems) if you construct a right triangle to inform your substitutions.



          Imagine a right triangle whose opposite (to $theta$) leg has length $2x$, adjacent leg has length $3$, and hypotenuse has length $sqrt{4x^2 + 9}$. From this I can deduce that $x = frac{3}{2}tantheta$, and therefore $dx = frac{3}{2}sec^2theta dtheta$. I can also determine that $sqrt{4x^2 + 9} = 3sectheta$. Substituting all this in gives:
          $$
          intfrac{x^3}{(4x^2+9)^frac{3}{2}} = int frac{left(frac{3}{2}tanthetaright)^3}{left(3secthetaright)^3} cdot frac{3}{2}sec^2theta dtheta = frac{3}{16}intfrac{tan^3theta}{sectheta}dtheta.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 5 at 22:23

























          answered Jan 5 at 22:00









          Austin MohrAustin Mohr

          20.1k35098




          20.1k35098























              0












              $begingroup$

              The answers provided so far already outline exactly how to approach your question with trigonometric substitutions. Here I will speak more broadly about the common trig substitutions. The general goal is to take some rational function of multiple terms and reduce it down to a single term. For instance, here you want to have $4x^2 + 9$ to be represented by a single term.



              This motivation of using trigonometry to reduce multiple terms to a single (or overall a much reduced amount) is quite common within integration. Although there are many different identities used, the majority are centred around the principle identity:



              begin{equation}
              sin^2(x) + cos^2(x) = 1
              end{equation}



              From here three core forms can be derived. The first is just a simple rearrangement:



              begin{equation}
              sin^2(x) + cos^2(x) = 1 rightarrow cos^2(x) = 1 - sin^2(x)
              end{equation}



              The next two are formed by dividing the principle identity through by $cos^2(x)$
              begin{equation}
              frac{1}{cos^2(x)}left(sin^2(x) + cos^2(x) right) = frac{1}{cos^2(x)} rightarrow tan^2(x) + 1 = sec^2(x)
              end{equation}



              Combining the three we have:




              1. $sec^2(x) = tan^2(x) + 1$

              2. $tan^2(x) = sec^2(x) - 1$

              3. $sin^2(x) = 1 - cos^2(x)$


              So here we two three different situations in which multiple terms can be reduced into a single term. These are (and to match the ordering of the list above)




              1. $a^2 + x^2$

              2. $x^2 - a^2$

              3. $a^2 - x^2$


              For your question you are working with (1).



              Edit - Sorry I should have said that not only are you looking to replace multiple terms with a single term but you are also often looking to get rid of a 'square root'. You will see in the list above that in all three cases the resultant term is a squared term and thus when taking the square root you get rid of it!






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                The answers provided so far already outline exactly how to approach your question with trigonometric substitutions. Here I will speak more broadly about the common trig substitutions. The general goal is to take some rational function of multiple terms and reduce it down to a single term. For instance, here you want to have $4x^2 + 9$ to be represented by a single term.



                This motivation of using trigonometry to reduce multiple terms to a single (or overall a much reduced amount) is quite common within integration. Although there are many different identities used, the majority are centred around the principle identity:



                begin{equation}
                sin^2(x) + cos^2(x) = 1
                end{equation}



                From here three core forms can be derived. The first is just a simple rearrangement:



                begin{equation}
                sin^2(x) + cos^2(x) = 1 rightarrow cos^2(x) = 1 - sin^2(x)
                end{equation}



                The next two are formed by dividing the principle identity through by $cos^2(x)$
                begin{equation}
                frac{1}{cos^2(x)}left(sin^2(x) + cos^2(x) right) = frac{1}{cos^2(x)} rightarrow tan^2(x) + 1 = sec^2(x)
                end{equation}



                Combining the three we have:




                1. $sec^2(x) = tan^2(x) + 1$

                2. $tan^2(x) = sec^2(x) - 1$

                3. $sin^2(x) = 1 - cos^2(x)$


                So here we two three different situations in which multiple terms can be reduced into a single term. These are (and to match the ordering of the list above)




                1. $a^2 + x^2$

                2. $x^2 - a^2$

                3. $a^2 - x^2$


                For your question you are working with (1).



                Edit - Sorry I should have said that not only are you looking to replace multiple terms with a single term but you are also often looking to get rid of a 'square root'. You will see in the list above that in all three cases the resultant term is a squared term and thus when taking the square root you get rid of it!






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The answers provided so far already outline exactly how to approach your question with trigonometric substitutions. Here I will speak more broadly about the common trig substitutions. The general goal is to take some rational function of multiple terms and reduce it down to a single term. For instance, here you want to have $4x^2 + 9$ to be represented by a single term.



                  This motivation of using trigonometry to reduce multiple terms to a single (or overall a much reduced amount) is quite common within integration. Although there are many different identities used, the majority are centred around the principle identity:



                  begin{equation}
                  sin^2(x) + cos^2(x) = 1
                  end{equation}



                  From here three core forms can be derived. The first is just a simple rearrangement:



                  begin{equation}
                  sin^2(x) + cos^2(x) = 1 rightarrow cos^2(x) = 1 - sin^2(x)
                  end{equation}



                  The next two are formed by dividing the principle identity through by $cos^2(x)$
                  begin{equation}
                  frac{1}{cos^2(x)}left(sin^2(x) + cos^2(x) right) = frac{1}{cos^2(x)} rightarrow tan^2(x) + 1 = sec^2(x)
                  end{equation}



                  Combining the three we have:




                  1. $sec^2(x) = tan^2(x) + 1$

                  2. $tan^2(x) = sec^2(x) - 1$

                  3. $sin^2(x) = 1 - cos^2(x)$


                  So here we two three different situations in which multiple terms can be reduced into a single term. These are (and to match the ordering of the list above)




                  1. $a^2 + x^2$

                  2. $x^2 - a^2$

                  3. $a^2 - x^2$


                  For your question you are working with (1).



                  Edit - Sorry I should have said that not only are you looking to replace multiple terms with a single term but you are also often looking to get rid of a 'square root'. You will see in the list above that in all three cases the resultant term is a squared term and thus when taking the square root you get rid of it!






                  share|cite|improve this answer











                  $endgroup$



                  The answers provided so far already outline exactly how to approach your question with trigonometric substitutions. Here I will speak more broadly about the common trig substitutions. The general goal is to take some rational function of multiple terms and reduce it down to a single term. For instance, here you want to have $4x^2 + 9$ to be represented by a single term.



                  This motivation of using trigonometry to reduce multiple terms to a single (or overall a much reduced amount) is quite common within integration. Although there are many different identities used, the majority are centred around the principle identity:



                  begin{equation}
                  sin^2(x) + cos^2(x) = 1
                  end{equation}



                  From here three core forms can be derived. The first is just a simple rearrangement:



                  begin{equation}
                  sin^2(x) + cos^2(x) = 1 rightarrow cos^2(x) = 1 - sin^2(x)
                  end{equation}



                  The next two are formed by dividing the principle identity through by $cos^2(x)$
                  begin{equation}
                  frac{1}{cos^2(x)}left(sin^2(x) + cos^2(x) right) = frac{1}{cos^2(x)} rightarrow tan^2(x) + 1 = sec^2(x)
                  end{equation}



                  Combining the three we have:




                  1. $sec^2(x) = tan^2(x) + 1$

                  2. $tan^2(x) = sec^2(x) - 1$

                  3. $sin^2(x) = 1 - cos^2(x)$


                  So here we two three different situations in which multiple terms can be reduced into a single term. These are (and to match the ordering of the list above)




                  1. $a^2 + x^2$

                  2. $x^2 - a^2$

                  3. $a^2 - x^2$


                  For your question you are working with (1).



                  Edit - Sorry I should have said that not only are you looking to replace multiple terms with a single term but you are also often looking to get rid of a 'square root'. You will see in the list above that in all three cases the resultant term is a squared term and thus when taking the square root you get rid of it!







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 5 at 23:01

























                  answered Jan 5 at 22:48









                  DavidGDavidG

                  2,022620




                  2,022620






























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