# Integration without Trig Sub

• Jul 11th 2009, 07:26 AM
latavee
Integration 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, 07:45 AM
CaptainBlack
Quote:

Originally Posted by latavee
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!

Split the integrand up $u=x^2$, and $dv=x\sqrt{x^2-9} \;dx$ then use integration by parts.

CB
• Jul 11th 2009, 08:22 AM
latavee
THanks!
AH! Never thought of that! Thank you!
• Jul 11th 2009, 08:37 AM
halbard
How about the substitution $u=\surd(x^2-9)$ or equivalently $x^2=u^2+9$? Then $x\,\mathrm dx=u\,\mathrm du$ and you end up with a polynomial in $u$ to integrate.
• Jul 11th 2009, 11:08 AM
HallsofIvy
I would do $\int x^3\sqrt{x^2- 9}dx$ by writing it as $\int \left(x^2\sqrt{x^2-9}l\right)\left(xdx\right)$ and using the substitution $u= x^2- 9$. Then du= 2xdx so $\frac{1}{2}du= xdx$ and $x^2= u+ 9$. That gives $\int x^3\sqrt{x^2- 9}dx= \frac{1}{2}\int (u+ 9)\sqrt{u}du$ $= \frac{1}{2}\int u^{3/2}+ 9u^{1/2} du$
• Jul 11th 2009, 11:31 AM
latavee
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!