Help! Integrate √(1+(1/(1-x)² - 1)²)?

Question

I have to turn this homework problem in by Monday! Please help! Hurry!

There IS an explicit expression for the indefinite integral. Don't say it can't be done.

Frank, the placement of the parathenses is the way I've got it. Your suggestion changes the problem.

I definitely need a closer look at smci's answer, so I've extended this question.

smci, you came close. The subsitution q = 1/(1-x)² - 1 seems fruitful, but had the integral been √(1-(1/(1-x)² - 1)²), it would have led to a simpler result, not far from what you've done. However, because it's √(1+(1/(1-x)² - 1)²) (see the + sign instead of the - sign), even with this substitution, it still leads to the same form of solution as suggested by Wombat below. Your effort was heroic, though, and I've spent quite a bit of time trying to figure out what went wrong.

Answer 1

When you see the standard forms 1/(1-x)² or √1/(1-x)²
you start thinking a trigonometric or hyperbolic substitution.
I think the most promising substitutions look like involving:
csc θ = √(1 + cot²θ)
cos θ = 1/√(1 + tan²θ)
sec θ = 1 + tan²θ

I = Integ √(1+ [1/(1-x)² - 1]²) dx

Then if we set:
[1/(1-x)² - 1] = cot θ

this reduces the integral
I to just Integ sec θ dx = Integ sec θ
but dx gets more complicated
Still I think that's the way to go.

1/(1-x)² - 1 = cot θ
-2dx/(1-x)³ = -csc²θ dθ
(-2/(1-x)) (1/(1-x)²) dx = -csc²θ dθ
(-2/(1-x)) (1/(1-x)² -1 +1) dx = -csc²θ dθ
(-2/(1-x)) (cot θ +1) dx = -csc²θ dθ

To eliminate 1/(1-x) factor, from above:
1/(1-x)² - 1 = cot θ
1/(1-x)² = 1 + cot θ
1/(1-x) = √(1 + cot θ)

Now can we can get dx by itself:
(-2/(1-x)) (cot θ +1) dx = -csc²θ dθ
-2√(1 + cot θ) (1 + cot θ) dx = -csc²θ dθ
2 (1 + cot θ)^(3/2) dx = csc²θ dθ
dx = 2 csc²θ (1 + cot θ)^(3/2) dθ

Finally the transformed integral is:
I = Integ √(1+ [1/(1-x)² - 1]²) dx
= Integ sec θ dx
= Integ sec θ * 2 csc²θ (1 + cot θ)^(3/2) dθ
= 2 Integ sec θ csc²θ (1 + cot θ)^(3/2) dθ

Want to eliminate the ugly (1 + cot θ)^(3/2) factor:
(1 + cot θ)^(3/2) = (1 + cos θ/sin θ)^(3/2)
= (sin θ/sin θ + cos θ/sin θ)^(3/2)
= ((sin θ + cos θ)/sin θ)^(3/2)
= ((sin θ + cos θ)/sin θ)^(3/2)

Then:
I = 2 Integ (1/cos θ) (1/sin²θ) (sin θ + cos θ)^(3/2) / (sin θ)^(3/2) dθ

This is a polynomial you can crank out one way or the other,
e.g. by transforming to a polynomial in one of these fns:
I tried: s = sin θ
ds = cos θ dθ
=> dθ = √1-s² ds

I = 2 Integ (1/√1-s²) 1/s² (s + √1-s²)^(3/2) / s^(3/2) √1-s² ds
= 2 Integ 1/s² (s + √1-s²)^(3/2) / s^(3/2) ds
doesn't look nice, so let's transform to q = cot θ instead:
sin θ = 1/√1+q²
cos θ = q/√1+q²
csc θ = √1+q²
sec θ = (√1+q²)/q
dq = -csc² θ dθ = -(q²+1)/q² dθ
dq = -(1 + 1/q²) dθ
dθ = -q²/(q²+1) dq

I = 2 Integ sec θ csc²θ (1 + cot θ)^(3/2) dθ
I = 2 Integ csc³ θ √cot θ dθ

or I = 2 Integ (√(1+q²) /q) (1+q²) q^(3/2) (-q²/(q²+1)) dq
I = -2 Integ q^(5/2)√(1+q²) dq but that goes nowhere

So: I = 2 Integ csc³ θ √cot θ dθ
Further substitution:
p = cot²θ
csc θ = √(1+p)
dp = 2cotθ (-csc²θ dθ)
dp = -2p(p+1) dθ
dθ = -dp/2p(p+1)

Hence:
I = 2 Integ csc³ θ √cot θ dθ
= 2 Integ (√(1+p))³ (p) (-dp/2p(p+1))
= - Integ √(1+p) dp

Yet another substitution gets us to the solution:
n = √(1+p)
dn = dp / 2√(1+p) = dp/2n
dp = 2n dn

I = - Integ √(1+p) dp
= - Integ n (2n dn)
= - Integ 2n² dn
= - 2/3 n³ + C
= -2/3 (1+p)^(3/2) + C
= -2/3 (1+ cot²θ )^(3/2) + C
= -2/3 (1+ [1/(1-x)² - 1]²)^(3/2) + C
SOLUTION!

where we applied the back-transformations:
n = √(1+p)
p = cot²θ
q = √p = cot θ = [1/(1-x)² - 1]

[This took me 2.5 hrs... I guess this one beats Matlab and Mathematica's heuristics]

Answer 2

It can't be done.
You need more parentheses'
sqrt((1+/(((1-x)^2) -1))^2) ?
If so then the integration is very complicated with 8 terms having sqrt and ln. You would need a CAS to solve by Monday.

FWIW Maple7 won't take your equation as written though Mathematica does.

Why are they doing this to you?

Answer 3

It can be done, but it's insane.

I plugged it into the intregrator at wolfram just to see what answer we're aiming for, and there's no way I could even approach that problem. The answer involves literally dozens of terms, many of them complex, as well as three other functions.

To see it for yourself, go to http://integrals.wolfram.com/index.jsp and input the following: Sqrt[1+(1/((1-x)^2)-1)^2]

Answer 4

It does exist, but it's complex, and It's extremely ugly, involving elliptic integral functions.
http://en.wikipedia.org/wiki/Elliptic_integral

Here's what Mathematica gives me
((Sqrt[1 + (-1 + (-1 + x)^-2)^2]* (-1 + x)*(1 - 4x + 10x^2 - 8x^3 + 2x^4 + 2i Sqrt[-1 - i]*(-1 + x)*Sqrt[-i + (2 + 2i)x - (1 + i)x^2]* Sqrt[I + (2 - 2i)*x - (1 - i)x^2]*EllipticE[ i*ArcSinh[Sqrt[-1 - i]*(-1 + x)], -i] - (2*(-1 + x)*Sqrt[-i + (2 + 2i)x - (1 + i)x^2]*Sqrt[ i + (2 - 2i)*x - (1 - i)x^2]*EllipticF[ i*ArcSinh[Sqrt[-1 - i]* (-1 + x)], -i])/ Sqrt[-1 - i]))/(1 - 4x + 10x^2 - 8x^3 + 2x^4))

I retrospect, there's probably some type of human error involved here.....

Answer 5

This function can not be integrated