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Homework Help: Trig Subsitution problem

  1. Mar 5, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{x^3}{\sqrt{x^2 + 9}}[/tex]


    2. Relevant equations

    [tex]x = 3\tan{\theta}[/tex]
    [tex]dx=3\sec^2{\theta}[/tex]


    3. The attempt at a solution

    [tex]27\int\tan^3{\theta}\sec{\theta}[/tex]
    [tex]27\int\tan{\theta}(\sec^2{\theta} - 1)\sec{\theta}[/tex]
    [tex]27\int(sec^3{\theta} - \sec{\theta})\tan{\theta}[/tex]
    [tex]27[\int\sec^3{\theta}\tan{\theta} - \int\sec{\theta}\tan{\theta}[/tex]
    [tex]27(\frac{1}{3}\sec^3{\theta} - \sec{\theta})[/tex]
    [tex]= 27[\frac{1}{3}(\frac{\sqrt{x^2 + 9}}{3})^3 - \frac{\sqrt{x^2 + 9}}{3}] + C[/tex]

    The correct answer is as follows:

    [tex]\frac{1}{3}(x^2 - 18)\sqrt{x^2 + 9}[/tex]


    Any ideas?
     
    Last edited: Mar 5, 2007
  2. jcsd
  3. Mar 5, 2007 #2
    Both are correct. You just haven't simplified your answer to the form given in the book.

    Btw, Welcome to PF! :)
     
  4. Mar 5, 2007 #3
    Thanks for the welcoming. Love the forums.

    I'm trying to simplify it but end up with something different. Where does the -18 come from? I've done the following:

    [tex]9(\frac{\sqrt{x^2 + 9}}{3})^3 - \frac{27\sqrt{x^2 + 9}}{3}[/tex]
    [tex]9[\frac{(x^2 + 9)\sqrt{x^2 + 9}}{27}] - 9\sqrt{x^2 + 9}[/tex]
    [tex]\frac{x^2 + 9\sqrt{x^2 + 9}}{3} - 9\sqrt{x^2 + 9}[/tex]
    [tex]\frac{[(x^2 + 9) - 9]\sqrt{x^2 + 9}}{3}[/tex]
    [tex]\frac{x^2\sqrt{x^2 + 9}}{3}[/tex]

    Thanks.
     
    Last edited: Mar 5, 2007
  5. Mar 6, 2007 #4
    [tex]27\left[\frac{\sqrt{x^2+9}}{3}\frac{(x^2+9)}{27} - \frac{\sqrt{x^2+9}}{3}\right]
    [/tex]

    [tex] = 27\left[\frac{\sqrt{x^2+9}}{3}\left(\frac{(x^2+9)}{27} - 1\right)\right][/tex]
     
    Last edited: Mar 6, 2007
  6. Mar 6, 2007 #5

    Gib Z

    User Avatar
    Homework Helper

    The first bit of tex you typed, you didn't have a dx stuck on the end. Ill sound pedantic, but you really should not forget them.
     
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