# Homework Help: Help with indefinite integral

1. Jul 5, 2010

### nlsherrill

1. The problem statement, all variables and given/known data
The problem reads(from the 4th edition of Stewart Calculus "Concepts and Contexts" pg. 394 #17):

Evaluate the integral.

the indefinite integral of dx/x^2*sqrt(4-x^2)

so this would read out as "the indefinite integral of dx over x squared times the square root of 4 minus x squared"

2. Relevant equations

3. The attempt at a solution

I tried everything I know. I have been stuck on this problem for over 2 hours now and I just cant see what to do. The back of the book says the answer is

-sqrt(4-x^2)/4x

the frustrating thing is that using substitution I am getting to -sqrt(4-x^2)/x so I feel like I'm pretty close, or just have some careless mistake somewhere I am missing.

2. Jul 5, 2010

### rock.freak667

Try a trig substitution x=2sint

3. Jul 5, 2010

### nlsherrill

I did already... but I guess I did it wrong.

After the substitution I get to 1/4*integral sin^2t then do I go to the half-angle formula for sin^2t? By doing that I get 1/4*integral(1/2)(1-cos2t), then integrate to get 1/8*(t-.5sin2t).

after that I solved for t using the trig substitution above to get t=arcsin(x/2). I then plug that into 1/8*(t-.5sin2t) to get 1/8*(arcsin(x/2)-.5sin2(arcsin(x/2), which I know is not right :/.

4. Jul 5, 2010

### Dick

To do the integral of 1/sin(t)^2 you are making it harder than it needs to be. You don't need any half angle formulas. What's the derivative of cot(t)? BTW write something like what you mean as 1/(4*sin^2(t)). Using parentheses will save you spelling it out in words.

Last edited: Jul 5, 2010
5. Jul 5, 2010

### nlsherrill

well I see that 1/sin(t)^2 =csc(t)^2, and the antiderivative of that is -cot(t)+c so i get all the way too

-1/4*cot(t) where t=arcsin(x/2). Am I on the right track?

6. Jul 5, 2010

### Dick

If you can figure out why -1/4*cot(t) where t=arcsin(x/2) gives you -sqrt(4-x^2)/(4x) then you are all the way there. Yes, you are on the right track.

Last edited: Jul 5, 2010
7. Jul 5, 2010

### nlsherrill

alright thanks a lot. Its really late and I'd rather not spend 2 more hours stuck doing the wrong thing. Your(and others) assistance is much appreciated.

8. Jul 5, 2010

### Dick

You aren't doing the wrong thing! Probably wouldn't take 2 hours to figure out why cot(arcsin(x/2))=sqrt(1-x^2/4)/(x/2) but you are probably right. It will take you even less time in the morning.

9. Jul 5, 2010

### nlsherrill

haha I know I'm not doing the wrong thing. I guess I meant to say im glad you told me I was on the right track because I wouldn't want to spend more time on this question tonight. 6 a.m. is calling my name for some calculus.

10. Jul 5, 2010

### Dick

Sleep agreeably knowing you'll get it the first thing tomorrow. You are SO CLOSE. Nite.

11. Jul 6, 2010

### Susanne217

Firstly it would be a whole lot easier to read if you wrote it in Latex like this:

$$\int \frac{dx}{x^2 \cdot \sqrt{4-x^2}}$$

And I can give you the hint that in the chapter dealing with these types of integrals there is a hint on howto solve this integral!

Last edited: Jul 6, 2010
12. Jul 6, 2010

### nlsherrill

Cool! Where's the 'hint' located in the chapter?

Yes I can see that Latex is much easier to read. I guess I'll work on learning to use Latex.

13. Jul 6, 2010

### hunt_mat

Here is an alternative:
$$\int\frac{dx}{x^{2}\sqrt{4-x^{2}}}$$
Use $$x=2\cos u$$ to transform the integral to:
$$-\frac{1}{4}\int\frac{du}{\cos^{2}u}=-\frac{1}{4}\int \sec^{2}udu$$
What well known function has derivative sec^{2}x?
when you see a factor of $$\sqrt{a^{2}-x^{2}}$$ your first thought should be trig substitution.