# Integration by trig substitution

• June 3rd 2008, 09:07 PM
Craka
Integration 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.
• June 3rd 2008, 09:24 PM
o_O
$x = 3\sin u$
$dx = 3\cos u \: du$

$\int {\color{red}x^{3}} {\color{blue}\sqrt{9 -x^{2}}} {\color{magenta}dx}$
$= \int {\color{red}\left(3\sin u\right)^{3}} {\color{blue}\sqrt{9 - 9\cos^{2} u}} \cdot {\color{magenta}3 \cos u \: du}$
$= \int 27 \sin^{3} u \cdot 3 \sqrt{1 - \cos^{2} u} \cdot 3\cos u \: du$
$= 243 \int \sin^{4} u \cdot \cos u \: du$

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 ;)
• June 3rd 2008, 10:27 PM
Craka
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?
• June 4th 2008, 01:19 PM
ebaines
Quote:

Originally Posted by Craka
How do you guys use the correct mathmatic symbols in the post?

See the sticky in the LaTex help forum.