Trigonometric substitution integral Trying to work around this with trig substitution, but end up with messy powers on sines and cosines... It should be simple use of trigonometric properties, but I seem to be tripping somewhere.
$$\int x^5\sqrt{x^2+4}dx $$
Thanks.

You don't even need a trigonometric substitution. Let $u^2=x^2+4$. Then $2u\,du=2x\,dx$, so $x\,dx=u\,du$.
Rewrite $x^5\sqrt{x^2+4}\,dx$ as $x^4\sqrt{x^2+4}\,x\,dx$, and substitute. Note that $x^2=u^2-4$, so $x^4=(u^2-4)^2$. Finishing the substitution, we get $(u^2-4)^2 u^2\,du$. Expand and integrate. No trig!
