Definate integral involving trig sub.

[amb1989]

Homework Statement

Take the integral of (sqrt(x^2-1)/X) bounded from 2 to 1

Homework Equations

The Attempt at a Solution

(o is supposed to stand for theta)

I used the substitution that said x is equal to asec(o) and solved for sec(o). From that I got that x = sec(o). So I set up the triangle I said the adj side was one, the oppisite is sqrt(x^2-1) and the hypt is x. I also took the derivative sec(o) dx and that equals sec(o)tan(o).

The next thing I did was substitute sec(o) in for all of the x values. So I got (sqrt(sec^2(o)-1)/sec(o)) * 1/sec(o)tan(o). The 1/sec(o)tan(o) piece takes place of the dx in the integral. So the sec^2(o) converts to tan^2(o) and the sqrt of that is just tan(o). So now it looks like tan(o)/sec(o) * 1/tan(o)sec(o). The tan(o) cancels and I get sec^2(o) in the denominator. 1/sec^2(o). I converted sec^2(o) into 1/cos^2(o). Using my triangle I replaced that with (1/x)^2. I took the integral of this and got 1/2ln(x^2). Plugging in my values I get 1/2ln(4)-1/2ln(1).

I factored out the 1/2 and combined the ln's and got 1/2(ln4/1) or just 1/2(ln4). This is wrong and I need help haha. PLEASE.

 

[Galadirith]

Hi amb1989,

That was an interesting way of trying to solve the problem, using the idea of the trig functions in triangles, never tried that myself. So your wanting to evaluate:

<br /> \int_1^2 \frac{\sqrt{x^2-1}}{x} \ dx<br />

Ill assume that the lower bound was in fact 1 and not 2 as the order you wrote them suggests.
So you wanted to use the substitution x = sec(θ). You also found the derivative which is necessary:

<br /> \frac{dx}{d\theta} \ = \ sec(\theta)tan(\theta) \ (i)<br />

You correctly wanted to substitute the x's in the integral with sec(θ) which leaves you with:

Hi amb1989,

That was an interesting way of trying to solve the problem, using the idea of the trig functions in triangles, never tried that myself. So your wanting to evaluate:

<br /> \int_1^2 \frac{\sqrt{sec^2(\theta)-1}}{sec(\theta)} \ dx<br />

however here's where you make the mistake, you said:

<br /> dx = \frac{1}{sec(\theta)tan(\theta)}d\theta<br />

which hopefully you can see isn't correct, looking at equation (i), it should be evident that it is infact:

<br /> dx = sec(\theta)tan(\theta)d\theta<br />

so the correct substituted integral should be:

<br /> \int_1^2 \frac{\sqrt{sec^2(\theta)-1}}{sec(\theta)} sec(\theta)tan(\theta) \ d\theta<br />

Now have a go from there, you should have to use triangle that you set up at all, just simplify the integral above and have a think :D