Integrating Trig Substitution: Simplifying the Solution

[teneleven]

Homework Statement

\int\frac{x^3}{\sqrt{x^2 + 9}}

Homework Equations

x = 3\tan{\theta}
dx=3\sec^2{\theta}

The Attempt at a Solution

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

The correct answer is as follows:

\frac{1}{3}(x^2 - 18)\sqrt{x^2 + 9}Any ideas?

 

[neutrino]

Both are correct. You just haven't simplified your answer to the form given in the book.

Btw, Welcome to PF! :)

 

[teneleven]

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:

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

Thanks.

 

[neutrino]

27\left[\frac{\sqrt{x^2+9}}{3}\frac{(x^2+9)}{27} - \frac{\sqrt{x^2+9}}{3}\right]<br />

= 27\left[\frac{\sqrt{x^2+9}}{3}\left(\frac{(x^2+9)}{27} - 1\right)\right]

 

[Gib Z]

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.