
trig substitution
the question was:
integral of squareroot(x^21) over x^2 dx
I substituted x for secx(t), and now I have tan^2(t)/sec(t) dt. I have no idea what to do from here, please help!
EDIT: Now I figured out it can be simplified to the integral of sin^2t/cost dt , or (1cos(2t))/2cost dt. Still can't figure it out or know which one to continue with...
Oh, also tried to do another problem,
integral of x^3 times squareroot(9+4x^2)dx
I think I'm supposed to substitute 3tan(t) in for x, but then I have the integral of 27tan^3(t) times 3 the squareroot of (1+4tan^2(t)) dt
and with the 4 in there I can't change it to sec^2(t), so what do I do?
Any help with either would be much appreciated, thank you!!

Doing some rearrangement:
$\displaystyle \int \frac{\sqrt{x^21}}{x^2}dx = \int (\frac{1}{\sqrt{x^21}}  \frac{1}{x^2 \sqrt{x^21}})dx$
The first part is: $\displaystyle \int \frac{1}{\sqrt{x^21}}dx = cosh^{1}x + C$
but I tried integration by parts and all sorts of substitutions but still couldn't get that second part.
However using a calculator I got the answer for the second part and worked back, but this process would not be obvious unless you knew the answer:
$\displaystyle \int \frac{1}{x^2 \sqrt{x^21}}dx = \int (\frac{1}{\sqrt{x^21}}  \frac{\sqrt{x^21}}{x^2})dx$ $\displaystyle = \int \frac{x^2  (x^21)}{x^2 \sqrt{x^21}}dx = \int \frac{\frac{x^2}{\sqrt{x^21}}\sqrt{x^21}}{x^2}dx$
$\displaystyle = \int \frac{x\frac{d}{dx}(\sqrt{x^21})\frac{d}{dx}(x)\sqrt{x^21}}{x^2}dx = \int \frac{d}{dx}(\frac{\sqrt{x^21}}{x})dx$ $\displaystyle = \frac{\sqrt{x^21}}{x} + c$
Hence the answer is: $\displaystyle \int = cosh^{1}x  \frac{\sqrt{x^21}}{x} + C$

as for the second part, put $\displaystyle x=\frac1t$ and you'll get a very easy integral to solve.