# 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 :\