# Homework Help: Two improper integrals

1. Jun 28, 2009

### clairez93

1. The problem statement, all variables and given/known data

Determine the divergence or convergence of the improper integral. Evaluate the integral if it converges.

1. $$\int^{4}_{2}$$$$\frac{1}{\sqrt{x^{2}-4}}$$ dx

2. $$\int^{2}_{0}$$$$\frac{1}{\sqrt[3]{x-1}}$$ dx

2. Relevant equations

3. The attempt at a solution

1.

t = x - 2
dt = dx

$$\int^{4}_{2}$$$$\frac{1}{\sqrt{x^{2}-4}}$$ dx
=$$\int^{2}_{0}$$$$\frac{1}{\sqrt{(t+2)^{2}-4}}$$ dx
=$$\int^{2}_{0}$$$$\frac{1}{\sqrt{t^{2}+4t}}$$ dx

$$\stackrel{lim}{x\rightarrow\infty}$$ $$\int^{2}_{N}$$$$\frac{1}{\sqrt{t^{2}+4t}}$$ dt

I'm not sure how to integrate $$\int^{2}_{N}$$$$\frac{1}{\sqrt{t^{2}+4t}}$$ dt from here.

2.

$$\int^{2}_{0}$$$$\frac{1}{\sqrt[3]{x-1}}$$ dx = $$\int^{1}_{0}$$$$\frac{1}{\sqrt[3]{x-1}}$$ + $$\int^{2}_{1}$$$$\frac{1}{\sqrt[3]{x-1}}$$

t = 1 - x
dt = -dx

$$\int^{1}_{0}$$$$\frac{1}{\sqrt[3]{x-1}}$$ = $$\int^{0}_{1}$$$$\frac{-1}{\sqrt[3]{-t}}$$ dt = $$\int^{1}_{0}$$$$\frac{1}{\sqrt[3]{-t}}$$ dt

$$\stackrel{lim}{x\rightarrow\infty}$$ $$\int^{1}_{N}$$$$-t^{-1/3}$$ dt
= $$\stackrel{lim}{x\rightarrow\infty}$$ -$$\frac{3}{4}(-t)^{4/3}$$$$|^{1}_{N}$$ = $$\stackrel{lim}{x\rightarrow\infty}$$ $$\left[\frac{3}{4} + \frac{3}{4}(-N)^{4/3}\right]$$ = $$\infty$$

So I got that this integral diverges, however, the book answers are:

1. $$ln (2 + \sqrt{3})$$
2. 0

Last edited: Jun 28, 2009
2. Jun 28, 2009

### clustro

3. Jun 28, 2009

### clairez93

Hmm, I see. I would like to know how this formula was derived though. Also do you see something wrong with #2?

4. Jun 28, 2009

### Dick

For the first one you do a trig substitution, like x=2*sec(t). For the second one the "improper" part of your integrand is it's divergence at x=1. You should be letting x->1 in both halves of the integral, not x->infinity.

Last edited: Jun 28, 2009
5. Jun 29, 2009

### g_edgar

In the first one, $$\lim_{N \to 0}$$ not $$\infty$$

6. Jun 29, 2009

### clairez93

But I set t = x - 1 specifially to make the discontinuity at 0; with t the problem occurs at 0, then don't we take the limit to infinity?

7. Jun 29, 2009

### Dick

Sure, in terms of t the integrand is (-t)^(-1/3) so the discontinuity is at 0. So for example integrate it from epsilon to 1 and let epsilon approach 0. Why would you send anything to infinity? BTW the antiderivative of (-t)^(-1/3) isn't (-3/4)*(-t)^(4/3). What's the correct power?

8. Jun 29, 2009

### clairez93

ah I just flipped through my textbook and you are right. I took the limit to 0 and got the right answer. the first I still am not sure how to integrate, there doesn't seem to be a trig substitution that works

9. Jun 29, 2009

### clairez93

the power you asked for is 2/3

10. Jun 29, 2009

### Dick

Your variable changes aren't really helping you. You are just moving the discontinuity around. To do the first integral stick with the form 1/sqrt(x^2-4) and substitute x=2*sec(t).

11. Jun 29, 2009

### Dick

That's better.

12. Jun 29, 2009

### clairez93

I don't understand how you came up with 2 sec t

13. Jun 29, 2009

### Dick

If x=2*sec(t) then x^2-4 becomes 4*sec(t)^2-4=4(sec(t)^2-1). sec(t)^2-1=tan(t)^2. Now the quantity inside the square root is a perfect square.

14. Jun 30, 2009

### clairez93

I tried doing the substitution but then got stuck again. This is what I did:

$$x = 2 sec (t)$$
$$dt = 2 sec(t) tan (t) dx$$
$$t = sec^{-1} (x/2)$$

$$\int^{\pi/3}_{0}$$$$\frac{1}{\sqrt{(2 sec(t)^{2} - 4} - 4}(2 sec(t) tan (t)}$$$$dt$$

= $$\int^{\pi/3}_{0}$$$$\frac{1}{\sqrt{4(sec^{2}t -1)}(2 sec(t) tan (t)}$$$$dt$$

= $$\int^{\pi/3}_{0}$$$$\frac{1}{\sqrt{4 tan^{2})t)}(2 sec(t) tan (t)}$$$$dt$$

= $$\int^{\pi/3}_{0}$$$$\frac{1}{2 tan(t)(2 sec(t) tan (t)}$$$$dt$$

= $$\int^{\pi/3}_{0}$$$$\frac{1}{2 tan^{2}(t)(2 sec(t))}$$$$dt$$

However, I still am not sure how to integrate this expression.

15. Jun 30, 2009

### Dick

If x=2*sec(t) then dx=2*sec(t)*tan(t)*dt. NOT dt=2*sec(t)*tan(t)*dx. The tan's will cancel and you should be left with just the integral of sec(t).

16. Jun 30, 2009

### Дьявол

Obviously you did something wrong.
$$x=2sec(t)$$

$$dx=2tan(t)sec(t)dt$$

Now,
$$\int^{\pi/3}_{0}\frac{1}{\sqrt{\frac{4}{cos^2(t)}-4}}*2tan(t)sec(t)dt=$$
$$=\int^{\pi/3}_{0}\frac{cos(t)}{\sqrt{4(1-cos^2(t))}}*2tan(t)sec(t)dt=$$
$$=\int^{\pi/3}_{0}\frac{cos(t)}{2sin(t)}*2tan(t)sec(t)dt=$$

Do you know how to continue from now on?

You got here cotan(t)*tan(t)=1 and its pretty easy to solve it.

Regards.

17. Jun 30, 2009

### clairez93

Finally solved it! Thank you! I would have never thought to use a trig substitution like that. Thanks.