Trying to do the indefinate integral of x^3 * (9-x^2)^1/2 by using the substitution of x = 3 sin u I get 81* integral[ (sin^3 u)(cos u)] but again I appear to be wrong.

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- Jun 3rd 2008, 09:07 PMCrakaIntegration by trig substitution
Trying to do the indefinate integral of x^3 * (9-x^2)^1/2 by using the substitution of x = 3 sin u I get 81* integral[ (sin^3 u)(cos u)] but again I appear to be wrong.

- Jun 3rd 2008, 09:24 PMo_O
$\displaystyle x = 3\sin u $

$\displaystyle dx = 3\cos u \: du$

$\displaystyle \int {\color{red}x^{3}} {\color{blue}\sqrt{9 -x^{2}}} {\color{magenta}dx}$

$\displaystyle = \int {\color{red}\left(3\sin u\right)^{3}} {\color{blue}\sqrt{9 - 9\cos^{2} u}} \cdot {\color{magenta}3 \cos u \: du}$

$\displaystyle = \int 27 \sin^{3} u \cdot 3 \sqrt{1 - \cos^{2} u} \cdot 3\cos u \: du$

$\displaystyle = 243 \int \sin^{4} u \cdot \cos u \: du$

I think you forgot to make the sub for dx. This last integral should be straightforward. Make the substitution s = sin u and you should be on your way ;) - Jun 3rd 2008, 10:27 PMCraka
Thanks for that you are right I had forgot to sub in the dx value. How do you guys use the correct mathmatic symbols in the post?

- Jun 4th 2008, 01:19 PMebaines