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!
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$