Simple Trigonometric Substitution Problem

[Sideline]

Homework Statement

\int{\frac{x^{3}}{\sqrt{4 - x^{2}}}}

NOTE: The use of "0" is theta. I couldn't figure out how to insert one :\

Homework Equations

The trig identity sin^{2}0 = 1 - cos^{2}0.

The Attempt at a Solution

I thought I completed the problem fine, but I realized WolframAlpha has a different (albeit slightly different) answer than mine. Can't figure out where I went wrong:

Because the square root is of the form a^{2} - u^{2}, the square root goes at the base, x is the height, and a is the hypotenuse:

u = x = 2sin0
dx = 2cos0d0
\sqrt{4 - x^{2}} = 2cos0

Therefore, the new integral is:
\int{\frac{(2sin0)^{3}2cos0d0}{2cos0}

Clean it up, and we have:
4 \times \int{sin^{3}0d0}

I split it up into sin squared, and sin, and then used the trig identity listed above to convert it into this:
4 \times \int{(1 - cos^{2}0)}sin0d0}

Multiply it out and separate the integrals:
4 \times \int{sin0d0} + 4 \times \int{(-sin0)(cos^{2}0)d0}

Both integrals are now simple to integrate; the first is -cos0, the second has u = cos, du = -sin:
4 \times-cos0 + 4 \times \frac{cos^{3}0}{3}

Replace cos0 with its corresponding values:
4 \times -\frac{\sqrt{4 - x^{2}}}{2} + 4 \times \frac{\sqrt{4 - x^{2}}^{3}}{3 \times 2} + C

Look about right? This (well, with reduced fractions, obviously) is what I have. No clue where I might have gone wrong :\

Thanks in advance!

 

[vela]

Looks basically right, but you messed up the factors of two in two places. First, when you cleaned up the integrand after the trig substitution, and second, when you substituted for cosine in the cos^3 term.

By the way, in LaTeX, \theta will get you the character you want.