How do I use trig substitution to solve this integral?

[p53ud0 dr34m5]

im hoping i worked this out right; its long:
\int x(81-x^2)^{5/2}dx
the integral contains a^2-x^2, so i set x=asin\theta. that would make x=9sin\theta and dx=9cos\theta d\theta:
\int 9sin\theta(81-81sin^2\theta)^{5/2}9cos\theta d\theta = \int 9sin\theta[81(1-sin^2\theta)]^{5/2}9cos\theta d\theta
the integral now contains 1-sin^2\theta =cos^2\theta
\int 9sin\theta(9cos\theta)^59cos\theta d\theta
i used u-sub by setting u=9cos\theta and du=-9sin\theta d\theta.
-\int u^6 du=-\frac{u^7}{7}+C
i plugged my u back in:
-\frac{(9cos\theta)^7}{7}+C
then, i drew my little triangle.
cos\theta=\frac{\sqrt{81-x^2}}{9}
i plugged that into the cos\theta and simplified and came out with:
-\frac{(\sqrt{81-x^2})^7}{7}+C
that's my answer

 

[Muzza]

You could also try the substitution u = 81 - x^2, which should be significantly easier.

 

[Theelectricchild]

I punched it into mathematica and it produced the same as yours, save for a 7/2 exponent rather than 7.

 

[p53ud0 dr34m5]

7/2 exponent is like my square root to the 7th
-\frac{(\sqrt{81-x^2})^2}{7}=-\frac{(81-x^2)^{7/2)}{7}
also, we had to use trig substitution, so that left out the easy 81-x^2 sub

 

[Hippo]

im hoping i worked this out right; its long:
\int x(81-x^2)^{5/2}dx

It doesn't have to be.

You shouldn't have to use a trig or u substitution. Inspection isn't too hard with something like this.

\frac{(2)}{(-2)(7)}\int (-2x)(\frac{7}{2})(81-x^2)^{5/2}dx
= -\frac{1}{7} (\sqrt {81-x^2})^7

 

[Theelectricchild]

*Cough* Table *Cough*

Excuse me... something in my throat--- tables are great to have around so you barely have to think.

 

[Hippo]

Guess having someone else do it is the easiest way.