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## Homework Statement

[tex]\int{\frac{x^{3}}{\sqrt{4 - x^{2}}}}[/tex]

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

## Homework Equations

The trig identity sin[tex]^{2}[/tex]0 = 1 - cos[tex]^{2}[/tex]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[tex]^{2}[/tex] - u[tex]^{2}[/tex], the square root goes at the base, x is the height, and a is the hypotenuse:

u = x = 2sin0

dx = 2cos0d0

[tex]\sqrt{4 - x^{2}}[/tex] = 2cos0

Therefore, the new integral is:

[tex]\int{\frac{(2sin0)^{3}2cos0d0}{2cos0}[/tex]

Clean it up, and we have:

[tex]4 \times \int{sin^{3}0d0}[/tex]

I split it up into sin squared, and sin, and then used the trig identity listed above to convert it into this:

[tex]4 \times \int{(1 - cos^{2}0)}sin0d0}[/tex]

Multiply it out and separate the integrals:

[tex]4 \times \int{sin0d0} + 4 \times \int{(-sin0)(cos^{2}0)d0}[/tex]

Both integrals are now simple to integrate; the first is -cos0, the second has u = cos, du = -sin:

[tex]4 \times-cos0 + 4 \times \frac{cos^{3}0}{3}[/tex]

Replace cos0 with its corresponding values:

[tex]4 \times -\frac{\sqrt{4 - x^{2}}}{2} + 4 \times \frac{\sqrt{4 - x^{2}}^{3}}{3 \times 2} + C[/tex]

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

Thanks in advance!

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