trig substitution

**ilovecalculus** · October 20th 2009, 06:54 PM

the question was:
integral of squareroot(x^2-1) 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 (1-cos(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!!

**calum** · October 21st 2009, 03:01 AM

Doing some rearrangement:

$\int \frac{\sqrt{x^2-1}}{x^2}dx = \int (\frac{1}{\sqrt{x^2-1}} - \frac{1}{x^2 \sqrt{x^2-1}})dx$

The first part is: $\int \frac{1}{\sqrt{x^2-1}}dx = cosh^{-1}x + C$

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:

$\int \frac{1}{x^2 \sqrt{x^2-1}}dx = \int (\frac{1}{\sqrt{x^2-1}} - \frac{\sqrt{x^2-1}}{x^2})dx$ $= \int \frac{x^2 - (x^2-1)}{x^2 \sqrt{x^2-1}}dx = \int \frac{\frac{x^2}{\sqrt{x^2-1}}-\sqrt{x^2-1}}{x^2}dx$

$= \int \frac{x\frac{d}{dx}(\sqrt{x^2-1})-\frac{d}{dx}(x)\sqrt{x^2-1}}{x^2}dx = \int \frac{d}{dx}(\frac{\sqrt{x^2-1}}{x})dx$ $= \frac{\sqrt{x^2-1}}{x} + c$

Hence the answer is: $\int = cosh^{-1}x - \frac{\sqrt{x^2-1}}{x} + C$

**Krizalid** · October 21st 2009, 03:23 AM

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

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