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]

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

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

 

[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.