Are these the same?

[grapz]

( 1 - x^2 ) ^2

(x^2 - 1)^2

when i expand this i get the same value. What am i doing wrong because they can't be the same right?

But i run into problems when trying to take the integral of this [ (x^2 + 1 ) ^2 / ( x^2 - 1) ^2] ^ 1/2 which gives me a different integral than this [(x^2 + 1)^2 / ( 1-x^2) ^2]^1/2


thanks

 

[pka]

Are these two the same \(\displaystyle \left|{a-b}\right|=\left|{b-a}\right|\)?
We know that \(\displaystyle a^2=b^2\) if and only if \(\displaystyle \left|a\right|=\left|b\right|\).

 

[stapel]

( 1 - x^2 ) ^2

(x^2 - 1)^2

Since x[sup:1ql3z1e1]2[/sup:1ql3z1e1] - 1 = -1(1 - x[sup:1ql3z1e1]2[/sup:1ql3z1e1]), and since squaring gets rid of the -1, these are the same.

But i run into problems when trying to take the integral of this [ (x^2 + 1 ) ^2 / ( x^2 - 1) ^2] ^ 1/2 which gives me a different integral than this [(x^2 + 1)^2 / ( 1-x^2) ^2]^1/2

What two integrals do you get? What were your steps? (The Integrator returns the same integral for both, as one would expect.)

Please be complete. Thank you!

Eliz.

 

[grapz]

The problem was finding the arc length of a certain graph.
I probably made a mistake somewhere but i can't find it
y = ln(1-x^2) from 0--> 1/2
y' = - 2x / (1 - x^2)

Using the formula L = int from 0--> 1/2 ( 1 + (dy/dx)^2 ) ^1/2

Subsitutating y' = -2x/(1-x^2) for dy/dx and simplyfing:

i get 4x^2 /(x^4 - 2x^2 + 1) + (x^4 - 2x^2 + 1)/ (x^4 - 2x^2 + 1)

which equals to [(x^2+1)^2 / (x^2-1)^2 ] ^1/2 = (x^2 + 1) / (x^2-1)

So its the integral of (x^2 + 1) / (x^2 -1)

Then i do x^2 +1 - 1 + 1 / ( x^2 - 1) = (x^2 - 1) / (x^2 - 1) + 2/(x^2-1)

which equals to 1 + 2/(x^2-1) . Using partial fractions, the integral is

x + ln(x-1) - ln(x+1). If i do the definite integral from 0--> 1/2. I get 1/2 - ln3/2

Now the problem arises because if i did the integral of (x^2+1) / (1-x^2) the integral is actually : -x + ln(1+x) - ln(1-x). The definite integral here is ln3 - 1/2

 

[stapel]

I think you dropped some bits along the way, including the square root...? The integral should be:

. . . . .\(\displaystyle \frac{(x^2\, -\, 1)\sqrt{\frac{(x^2\, +\, 1)^2}{(x^2\, -\, 1)^2}}\left(x\, +\, \ln{(x\, -\, 1)}\, -\, \ln{(x\, +\, 1)}\right)}{x^2\, +\, 1}\)

This may be one of those times when the absolute-value bars are fairly important...?

Eliz.

 

[pka]

This becomes \(\displaystyle \sqrt {1 + \left( {\frac{{ - 2x}}{{1 - x^2 }}} \right)^2 } = \sqrt {\frac{{1 - 2x^2 + x^4 + 4x^2 }}{{\left( {1 - x^2 } \right)^2 }}} = \sqrt {\frac{{\left( {1 + x^2 } \right)^2 }}{{\left( {1 - x^2 } \right)^2 }}}\)
But \(\displaystyle x \in \left[ {0,\frac{1}{2}} \right]\quad \Rightarrow \quad \sqrt {\frac{{\left( {1 + x^2 } \right)^2 }}{{\left( {1 - x^2 } \right)^2 }}} = \frac{{\left( {1 + x^2 } \right)}}{{\left( {1 - x^2 } \right)}}\)