How would I work the integral:

x^3 * sqrt(x^2 -9) dx

without using trig substitution? My professor said that it is a way, but I don't see it!

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- Jul 11th 2009, 06:26 AMlataveeIntegration without Trig Sub
How would I work the integral:

x^3 * sqrt(x^2 -9) dx

without using trig substitution? My professor said that it is a way, but I don't see it! - Jul 11th 2009, 06:45 AMCaptainBlack
- Jul 11th 2009, 07:22 AMlataveeTHanks!
AH! Never thought of that! Thank you!

- Jul 11th 2009, 07:37 AMhalbard
How about the substitution $\displaystyle u=\surd(x^2-9)$ or equivalently $\displaystyle x^2=u^2+9$? Then $\displaystyle x\,\mathrm dx=u\,\mathrm du$ and you end up with a polynomial in $\displaystyle u$ to integrate.

- Jul 11th 2009, 10:08 AMHallsofIvy
I would do $\displaystyle \int x^3\sqrt{x^2- 9}dx$ by writing it as $\displaystyle \int \left(x^2\sqrt{x^2-9}l\right)\left(xdx\right)$ and using the substitution $\displaystyle u= x^2- 9$. Then du= 2xdx so $\displaystyle \frac{1}{2}du= xdx$ and $\displaystyle x^2= u+ 9$. That gives $\displaystyle \int x^3\sqrt{x^2- 9}dx= \frac{1}{2}\int (u+ 9)\sqrt{u}du$$\displaystyle = \frac{1}{2}\int u^{3/2}+ 9u^{1/2} du$

- Jul 11th 2009, 10:31 AMlatavee
I'm going to try some of these ideas out! and I will see if I can come up with the answer that I had using trig substitution/or nearly the same! Thanks guys!