Trig Subsitution problem

  • Thread starter teneleven
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  • #1
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Homework Statement



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


Homework Equations



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


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:

Answers and Replies

  • #2
2,063
2
Both are correct. You just haven't simplified your answer to the form given in the book.

Btw, Welcome to PF! :)
 
  • #3
12
0
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:
  • #4
2,063
2
[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:
  • #5
Gib Z
Homework Helper
3,346
5
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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