another integration question!!

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katyt

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Sep 7, 2005
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Calculate the integral of

dx/1-x^2 both by partial fractions and by using a trigonometric substitution. Show that the results you get are the same.

I used partial fractions and got 1/2ln((1-x)/(1+x)) + C
And using a trig substitution I got 1/2ln((sinarcsinx+1)/(sinarcsinx-1)) + C
The trig substitution answer doesn't make sense to me, so I'm thinking I did something wrong. I used x=sin(theta) as the substitution. If someone could help me out I would really appreciate it!

Thanks again
Katy
 
Come on, Katy! What is sin(arcsin(x+1))? You'll kick yourself.
 
except that it's (sinarcsin(x)+1)/(sinarcsin(x)-1) , so I can't figure it out.
 
You missed the point.

What's sin(arcsin(frog))? How about sin(arcsin(Pittsburgh))?

Assuming 'frog' and 'Pittsburgh' are in the Domain of arcsin(x), anyway.
 
tkhunny said:
What's sin(arcsin(frog))? How about sin(arcsin(Pittsburgh))?

Assuming 'frog' and 'Pittsburgh' are in the Domain of arcsin(x), anyway.
Click to expand...
What you wrote implies that both

. . . . -1 <= frog <= 1

and

. . . . -1 <= Pittsburgh <= 1

You could publish a paper based on these findings, and make me a coauthor please.
 
Fine, now that Matt has taken us in yet another direction, I'll just spill it. :(

The DEFINITION of an inverse function includes f(f<sup>-1</sup>(x)) = x.

Interesting implications are:

sin(arcsin(x+1)) = x+1
sin(arcsin(x-1)) = x-1

NOW what do you think of your expression?
 
I knew that. I was laughing so hard I couldn't type.

What do you think, katyt? Are you getting it?
 
yea i get all of that, but i'm trying to prove that 1/2ln((1+x)/(1-x)) = 1/2ln((x+1)/(x-1)) and I'm thinking that they aren't equal because 1-x doesnt = x-1

so my question is, what did i do wrong?
 
Hi katyt,

I agree with your final answers, except you must remember that when we integrate 1/x we get ln |x|. That should help you see what's going on...
 
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