# Definate integral involving trig sub.

1. Oct 5, 2009

### amb1989

1. The problem statement, all variables and given/known data
Take the integral of (sqrt(x^2-1)/X) bounded from 2 to 1

2. Relevant equations

3. 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.

2. Oct 5, 2009

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:

$$\int_1^2 \frac{\sqrt{x^2-1}}{x} \ dx$$

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:

$$\frac{dx}{d\theta} \ = \ sec(\theta)tan(\theta) \ (i)$$

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:

$$\int_1^2 \frac{\sqrt{sec^2(\theta)-1}}{sec(\theta)} \ dx$$

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

$$dx = \frac{1}{sec(\theta)tan(\theta)}d\theta$$

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

$$dx = sec(\theta)tan(\theta)d\theta$$

so the correct substituted integral should be:

$$\int_1^2 \frac{\sqrt{sec^2(\theta)-1}}{sec(\theta)} sec(\theta)tan(\theta) \ d\theta$$

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