Help with a trig integral

OK I'm studying for a calculus exam and I ran across a problem where I thought I did everything right, but I can't know for sure because it's an even problem and I don't have an answer.

__Problem:__

$\displaystyle \int cos^5 (x) dx = $

__What I did:__

$\displaystyle \int cos^5 (x) dx = $

**Splitting into squares:**

$\displaystyle \int [cos^2 (x)]^2 cos (x) dx$

Using cos^2 + sin^2 = 1:

$\displaystyle \int [1 - sin^2 (x)]^2 cos (x) dx$

**Substituting:**

$\displaystyle u = \sin (x)$

$\displaystyle du = \cos (x) dx$

$\displaystyle \int (1 - u^2)^2 dx$

**Foiling:**

$\displaystyle \int (1 - 2u^2 - u^4) dx$

**Integrating:**

$\displaystyle u - \frac{2}{3}u^3 - \frac{1}{5}u^5$

**Unsubstituting:**

$\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x)$

__Answer:__

$\displaystyle sin (x) - \frac{2}{3}sin^3 (x) - \frac{1}{5}sin^5 (x) + C$

I typed the problem into Wolfram Alpha and it gave me this:

$\displaystyle \frac{5}{8}sin (x) - \frac{5}{48}sin (3x) - \frac{1}{80}sin (5x) + C$

This looks alot different than my answer, and they did it using something called a 'reduction formula', which seems like something I would never remember. Is my answer wrong, or did they just do it differently? Or is Wolfram Alpha not the best place to be checking my answers?

Thanks