How can I evaluate the indefinite integral of dx/x^2*sqrt(4-x^2)?

[nlsherrill]

Homework Statement

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"

Homework Equations

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 can't 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.

 

[rock.freak667]

Try a trig substitution x=2sint

 

[nlsherrill]

Try a trig substitution x=2sint

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 :/.

 

[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.

 

[nlsherrill]

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.

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?

 

[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.

 

[nlsherrill]

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.

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.

 

[Dick]

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.

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.

 

[nlsherrill]

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.

haha I know I'm not doing the wrong thing. I guess I meant to say I am 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.

 

[Dick]

haha I know I'm not doing the wrong thing. I guess I meant to say I am 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.

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

 

[Susanne217]

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

[tex]\int \frac{dx}{x^2 \cdot \sqrt{4-x^2}}[/tex]

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!

 

[nlsherrill]

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

[tex]\int \frac{dx}{x^2 \cdot \sqrt{4-x^2}}[/tex]

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!

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.

 

[hunt_mat]

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