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March 31, 2015

March 31, 2015

Posted by **Terry** on Wednesday, September 3, 2008 at 7:10am.

So I know that I am going to have to factor out the 4 and use the arcsec trig identity but I've having trouble getting to a viable u-substitution.

I've gotten to factoring like this so far:

√(x^2-4) = √[2^2(x^2/2^2 - 1)] = 2√((x/2)^2 - 1))

I could do

u = x/2

du = 1/2 dx

But that still leaves the x in front of the radical.

How can I solve this?

- Math integral -
**David Q**, Wednesday, September 3, 2008 at 4:25pmI'm extremely rusty at these, but I think it's possible you might be able to use the substitution x=2cosh(u), where cosh is the hyperbolic cosine. cosh²(u)-sinh²(u)=1, so sqrt(x²-4) = sqrt(4cosh²(u)-4) = 2sinh(u). dx/du=2sinh(u), so dx=2sinh(u)du, which cancels out with the square root in the denominator, leaving 2cosh(u) in the denominator to perform the integral on. I can't remember whether that's integrable or not: you'd need to look into that. Is that any help?

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