Can anyone help me figure out what I'm doing wrong:

$\displaystyle \int x^3\sqrt{9-x^2} dx$

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> Substitute:

> $\displaystyle x=3sin\theta$

> $\displaystyle dx=3cos\theta d\theta$

> $\displaystyle \theta = arcsin(x/3)$

> $\displaystyle cos(\theta)=cos(arcsin(x/3))= \frac{\sqrt{9-x^2}}3$

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=$\displaystyle \int (3sin\theta)^3\sqrt{9-(3sin\theta)^2}(3cos\theta) d\theta$

=$\displaystyle 3^4\int sin^3\theta\cos\theta\sqrt{3^2-3^2sin^2\theta} d\theta$

=$\displaystyle 3^4\int sin^3\theta\cos\theta\sqrt{3^2(1-sin^2\theta)} d\theta$

=$\displaystyle 3^4\int sin^3\theta\cos\theta\sqrt{3^2cos^2\theta} d\theta$

=$\displaystyle 3^5\int sin^3\theta\cos^2\theta d\theta$

=$\displaystyle 3^5\int sin\theta (1-cos^2\theta)\cos^2\theta d\theta$

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> Substitute:

> $\displaystyle u = cos\theta$

> $\displaystyle -du = sin\theta d\theta$

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=$\displaystyle -3^5\int (1-u^2)u^2 du$

=$\displaystyle -3^5\int (u^2-u^4) du$

=$\displaystyle -3^5(\frac13 u^3 - \frac15 u^5)+C$

=$\displaystyle -3^5(\frac13 cos^3\theta - \frac15 cos^5\theta)+C$

=$\displaystyle -3^5\left(\frac13 \left(\frac{\sqrt{9-x^2}}{3}\right)^3 - \frac15 \left(\frac{\sqrt{9-x^2}}{3}\right)^5\right)+C$

=$\displaystyle -3^5\left(\frac1{3^4} (9-x^2)^{3/2} - \frac1{5*3^5} (9-x^2)^{5/2}\right)+C$

=$\displaystyle -3 (9-x^2)^{3/2} + \frac15 (9-x^2)^{5/2}+C$

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I've done it several times and gotten the same answer, unfortunately it's an even problem so I can't check my book, but Function calculator says the answer should be:

$\displaystyle -\frac15 x^2(9-x^2)^{3/2}-\frac65(9-x^2)^{3/2}+C$