Straight Dope Message Board > Main Could someone please help me Integrate this equation?
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#1
07-07-2002, 10:57 PM
 Muad'Dib Guest Join Date: Jul 2000

First off, this is not part of a problem set, this is not direct homework. I am studying for a test tomorrow and am stuck on this problem.

Integrate( (1+tan)^(1/2) )

If someone could just tell me the next step, that would be enough.
#2
07-07-2002, 11:23 PM
 Muad'Dib Guest Join Date: Jul 2000
Ohhhh.... Wait a sec, I farked up. The equation is (1+tan^2)^(1/2). That is easily solved with a trignometric substitution.

However, I would still like to know if the other one can be solved.
#3
07-07-2002, 11:24 PM
 ultrafilter Guest Join Date: May 2001
That's sqrt(1 + tan(x)), right? I'm not sure what to do with that one. Maybe substitute for 1 + tan(x) and integrate by parts?

If, however, it were sqrt(1 + tan(x)2), then you'd just need to apply sin(x)2 + cos(x)2 = 1 (or an equivalent equation).
#4
07-07-2002, 11:29 PM
 ultrafilter Guest Join Date: May 2001
Can the first one be solved? Sure, it's a continuous function, and every continuous function is integrable.

Does it have an integral that you could write down? Well, maybe not. Actually, this one does--I'd give it to you, but I'm not sure where the parentheses go (I used The Integrator, and it gave me an ambiguous answer). But not every function has a closed form integral (i.e., one that can be finitely expressed without using an integral).
#5
07-08-2002, 06:00 AM
 ShadowWarrior Guest Join Date: Mar 2002
I would have thought Muad'dib could use presciense(sp?) to determine the answer from a couple of hours in the future, when he's worked it out
#6
07-08-2002, 06:17 AM
 Achernar Guest Join Date: Aug 1999
The integrand with the tan2 simplifies to Abs(Sec(x)), right? Anyway, the other, more insidious one you won't be seeing on a test. The Integrator that ultrafilter links to seems to work fine. In this case, you'd enter the input like this:

Sqrt[1+Tan[x]]

However, it gives you nasty results with hyperbolic arctangents and imaginary numbers, all of which we hope cancel for a real result.
#7
07-08-2002, 02:33 PM
 Chronos Charter Member Join Date: Jan 2000 Location: The Land of Cleves Posts: 50,790
Maple's result doesn't have any hyperbolics or explicitly complex numbers (although it does have a few log terms which might go imaginary), but it's still pretty ugly. If you really want to know, it's -1/4*(-sqrt(2*sqrt(2)+2)*ln(tan(x)+1+sqrt(2*sqrt(2)+2)*sqrt(1+tan(x))+sqrt(2))*sqrt(2*sqrt(2)-2)-4*arctan((2*sqrt(1+tan(x))+sqrt(2*sqrt(2)+2))/(2*sqrt(2)-2)^(1/2))+sqrt(2*sqrt(2)+2)*sqrt(2)*ln(tan(x)+1+sqrt(2*sqrt(2)+2)*sqrt(1+tan(x))+sqrt(2))*sqrt(2*sqrt(2)-2)+sqrt(2*sqrt(2)+2)*ln(-tan(x)-1+sqrt(2*sqrt(2)+2)*sqrt(1+tan(x))-sqrt(2))*sqrt(2*sqrt(2)-2)+4*arctan((-2*sqrt(1+tan(x))+sqrt(2*sqrt(2)+2))/(2*sqrt(2)-2)^(1/2))-sqrt(2*sqrt(2)+2)*sqrt(2)*ln(-tan(x)-1+sqrt(2*sqrt(2)+2)*sqrt(1+tan(x))-sqrt(2))*sqrt(2*sqrt(2)-2))/(2*sqrt(2)-2)^(1/2)

I sure as heck hope that wraps.
#8
07-08-2002, 02:35 PM
 ultrafilter Guest Join Date: May 2001
Chronos, did you try simplify(%) after you got that back?
#9
07-08-2002, 03:09 PM
 g8rguy Guest Join Date: Jun 2001
Simplify gives you a fat lot of nothing helpful. There's an overall factor of sqrt(2)-1 which you can pull out, and you can of course combine the logarithms, but it's still ugly. And I have no idea how it did the integral; I've been sitting over in my corner perioidically trying different substitutions and getting other things that I still can't integrate.
#10
07-08-2002, 03:12 PM
 ultrafilter Guest Join Date: May 2001
There are special commands to simplify using trigonometric rules and the like. I'll play with it once I get home, unless someone else tries it first. Of course, it's possible that there's nothing to be done with it, but I'd like to check.
#11
07-08-2002, 04:11 PM
 g8rguy Guest Join Date: Jun 2001
Eh, I'm pretty good with maple, but unless you can think of some relation for arctan(a) + arctan(b) (there may by one; I'm woeful with inverse trig functions), I'm stuck at A*(ln(big ugly thing) + B*(arctan(2nd big ugly thing) + arctan(3rd big ugly thing))), where A and B are some combination of sqrt(2)+1, sqrt(2)-1, and a few other things that I don't recall off the top of my head.
#12
07-08-2002, 07:27 PM
 ultrafilter Guest Join Date: May 2001
Nope, I can't get anything either. So yeah, sqrt(1 + tan(x)) does have a closed form integrand, but I'll be damned if I see how you'd get it.
#13
07-08-2002, 07:41 PM
 g8rguy Guest Join Date: Jun 2001
hmm... I tried making the simple change of variables u=-i*x, just because I like hyperbolic functions better, and something a bit simpler popped out of maple. Alas, I forgot to post it. And I'm at home and haven't bothered buying maple yet.

And even if I'm not hallucinating and we get something a bit nicer... I have no idea how i*sqrt(1+i*tanh(u)) is any easier. I move we leave this problem as an exercise for the reader. (I'm going to be using this phrase a lot in my career as a text book writer... "the proof is trivial and is left as an exercise to the reader." Very handy. )

I bet, though, that this involves some really obscure transformation like (1+ln(5u-3)) = (1+tan(x))^(17/5) or something obscene like that.

Oh, and I just thought I ought to point out that there's no reason to worry about getting imaginary numbers anywhere, since our integrand might be pure imaginary for certain values of x.
#14
07-08-2002, 10:06 PM
 Bryan Ekers Guest Join Date: Nov 2000
Chronos: -25% for not showing your work.
#15
07-09-2002, 08:03 PM
 g8rguy Guest Join Date: Jun 2001
Now proving once and for all that I have far far far too much time on my hands

Right, so I figured it out on the way home from the office (I have this irritating 20 minute walk, and so I got busy thinking...).

First, make the substitution u = tan(x) to give an integrand of
sqrt(1+u)/(1+u2).

Next, decompose the denominator in partial fractions:
1/(1+u2) = i/2*(1/(u+i) - 1/(u-i))

Make another change of variables to v = sqrt(1+u). At this point, our integrand is
i*v2*(1/(a2-v2) - 1/(b2-v2)
where a2 = 1+i and b2 = 1-i

Now notice that v2/(a2-v2) = a2/(a2-v2) - 1

Using this gives us something we can integrate:
i*[a2/(a2-v2) - b2/(b2-v2)]

Both terms are elementary integrals: Int(a2/(a2-v2)) = a*arctanh(v/a)

So we have, in the end
i*sqrt(1+i)*arctanh(v/sqrt(1+i)) - i*sqrt(1-i)*arctanh(v/sqrt(1-i))

and substituting in v = sqrt(1+u) = sqrt(1+tan(x)) gives the final answer:
i*sqrt(1+i)*arctanh(sqrt((1+tan(x))/(1+i))) - i*sqrt(1-i)*arctanh
(sqrt((1+tan(x))/(1-i)))

Spending a whole lot of time with arctanh(x) = 1/2*log((1+x)/(1-x)) and various other identities should eventually give the same answer as maple gets, since my answer does work out to be correct.

*not holding breath waiting for congratulations and awe from the SDMB, as this was a truly truly pathetic way to spend 30 minutes, but oh well*
#16
07-09-2002, 08:09 PM
 ultrafilter Guest Join Date: May 2001
Congratulations, awe, and all that.

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