Integration question (1 Viewer)

[ngogiathuan]

Here it is
integrate sqrt(4+x^2)
this question can be solved by letting x= 2 tan u
when we finish integrating this, the function will be left in term of sec u and tan u, so we need to swap them back to x. My question is, when we let x=2tan u, do we have to put restriction on u such that tan u and sec u are both >0 (if so, how we do that?), or do we just assume that both of them are >0 .
well thats it. Thanks in advance

 

[Slidey]

Are you asking about preventing division by zero, or asking about the restrictions possibly introduced by using inverse trig functions?

 

[ngogiathuan]

i mean when u do that integration sqrt (4+x^2), u let x=2tan u. After doing some fancy moves u are left with a function in sec u and tan u, and we need to swap back in term of x
Now sec^2 = tan^2+1 therefore sec = +- sqrt (tan^2+1), so how can i tell whether to take the +ve or -ve sign in this case
the question is sqrt (4+x^2), which mean when u let x = 2 tan u, u is not restricted since there are no restriction on x. so i cant tell if tan u >0 or <0
Its a bit clumsy but , in brief, should i assume that sec u tan u >0 or do sth?

 

[Affinity]

once you get it back into

[x(x^2+4)^(1/2)]/2 + 2 log(x + (x^2 + 4)^(1/2))

you can see that it doesn't really matter. hmm the tan substitution is not that bad, it's a one to one function. you can start with x = -2tan(u) and you will end up with the same thing.

 

[Slidey]

I imagine the reasons are a bit complex and tedious.

But consider that all trig functions are really various exponential and logarithmic functions (as defined by re^(i@)=rcis(@)=a+ib). As such, it's not as though sec suddenly stops working for the values between -1 and 1. As long as it still gives SOME sort of answer, things will hold together:

Let's see:

arcsec(x) = i*ln(sqrt(1/x^2 - 1)+1/x)
arcsec(1/2) = i*ln(sqrt(3)+2) = an imaginary number
arcsec(3/2) = i*ln(i*sqrt(5)/3+2/3) = most likely works out to be a real number... let's see: i*ln(e^(i@))=i*i*@=-@, where @=arctan(sqrt(5)/2) I think (it's @=arctan(y/x), right?). Yay it works!

As you can see, the only time it doesn't work is when x=0.

 

[ngogiathuan]

once you get it back into

[x(x^2+4)^(1/2)]/2 + 2 log(x + (x^2 + 4)^(1/2))

you can see that it doesn't really matter. hmm the tan substitution is not that bad, it's a one to one function. you can start with x = -2tan(u) and you will end up with the same thing.

I see what u mean hmmm i'll try x = -2tan(u) later
And to Slidey ur answer is way beyond my comprehension.
Thank u all for ur replies
Cheers

 

[Slidey]

Isn't the point here that inverse trig functions are missing part of their domain? For example, arcsec doesn't return a real value between -1 and 1? Wasn't your question "so does this answer for the integral hold between those values or not?" The answer to that is: yes, because whilst arcsec has no real answer for -1 < x < 1, arcsec does have a complex answer, and thus typical arithmetic and algebra still apply normally. Similarly, does tan exist or not at kpi/2?

tan(x)=(e^ix-e^-ix)/(i(e^ix+e^-ix))
tan(pi/2)=(e^ipi/2 - 1/e^ipi/2)/(ie^ipi/2 + i/e^ipi/2), now by e^(i*pi)=-1,
= i(-sqrt(e)+1/sqrt(e))/(sqrt(e)+1/sqrt(e)) - thus tan also exists at kpi/2 (as a complex number), so even in this case, there are no holes introduced when one uses a tan substitution.
Just so you're a little more convinced this expontential form for tan is correct:
tan(pi)=(-1+1)/(-i-i)=0 (which is correct)

 

[Iruka]

tan(x)=(e^ix-e^-ix)/(i(e^ix+e^-ix))
tan(pi/2)=(e^ipi/2 - 1/e^ipi/2)/(ie^ipi/2 + i/e^ipi/2), now by e^(i*pi)=-1,
= i(-sqrt(e)+1/sqrt(e))/(sqrt(e)+1/sqrt(e))

You might want to reconsider what you just did there.

Tan(z) most certainly isn't analytic at z=pi/2.

 

[ngogiathuan]

Isn't the point here that inverse trig functions are missing part of their domain? For example, arcsec doesn't return a real value between -1 and 1? Wasn't your question "so does this answer for the integral hold between those values or not?" The answer to that is: yes, because whilst arcsec has no real answer for -1 < x < 1, arcsec does have a complex answer, and thus typical arithmetic and algebra still apply normally. Similarly, does tan exist or not at kpi/2?

tan(x)=(e^ix-e^-ix)/(i(e^ix+e^-ix))
tan(pi/2)=(e^ipi/2 - 1/e^ipi/2)/(ie^ipi/2 + i/e^ipi/2), now by e^(i*pi)=-1,
= i(-sqrt(e)+1/sqrt(e))/(sqrt(e)+1/sqrt(e)) - thus tan also exists at kpi/2 (as a complex number), so even in this case, there are no holes introduced when one uses a tan substitution.
Just so you're a little more convinced this expontential form for tan is correct:
tan(pi)=(-1+1)/(-i-i)=0 (which is correct)

I think u have somehow misread my question. My original question was when i use the substitution x = 2 tan u, and need to find sec u in term of x, since sec^2 = tan^2 +1 hence sec = +- sqrt (tan^2+1), should i take the +ve or -ve sign in this case.
your reply is very much in detail and i appreciate for that.

 

[Slidey]

You might want to reconsider what you just did there.

Tan(z) most certainly isn't analytic at z=pi/2.

Why not? Was there a floor in my algebra?

Elaboration would be good.

 

[Slidey]

I think u have somehow misread my question. My original question was when i use the substitution x = 2 tan u, and need to find sec u in term of x, since sec^2 = tan^2 +1 hence sec = +- sqrt (tan^2+1), should i take the +ve or -ve sign in this case.
your reply is very much in detail and i appreciate for that.

So you're not interested in, for example, the fact that even though inverse sec is used to find the answer to the integral, we seemingly give an answer that applies to all values of x? What about x=1/2 for example?

Well, the answer is: both work. You only really use a tan substitution when you have a+bx^2 in your integrand (square root or not).

Besides, you don't choose x=tanu because of sec=sqrt(1+tan^2), you choose x=tanu because it squares to give tan^2(u). It isn't the only substitution, but any other substitution that works, such as integration by parts, exponential substitution, t-formulae, or x=-tanu, will (eventually) give the same answer.

Example: Integrate 1/(1+x^2) dx.

No square roots, right? But you still use x=tanu:
dx=sec^2(u)du
Answer: u+c = arctan(x)+c.
Likewise, we could also use x=-tanu,
But we don't pick these subs because they are the square root of something, we pick them because they SQUARE to give something (which is what we want). Very similar concepts, but the key difference is that squaring is a function, whilst, generally, square-rooting is a relation.

If you're asking whether or not the final answer to the integral you posed is positive or negative... I believe that depends on whether they mean principle square root (positive), or the square root defined by y^2=x (positive and negative). Most of the time people mean principle square root, since y^2=x is a relation not a function.

 

[Iruka]

e^ipi/2 + e^-ipi/2 = i-i = 0

You can't divide by zero, even in the complex plane.

 

[Slidey]

e^ipi/2 + e^-ipi/2 = i-i = 0

You can't divide by zero, even in the complex plane.

cis(pi/2)+cis(-pi/2)=cos(pi/2)=0.

Dammit. I'm a silly billy. I think I confused addition and multiplication. I.e. I thought e^(ipi/2)=e^(1/2)*e^(ipi).

So, Iruka, does that mean that the integral is undefined for x=kpi/2? I doubt it, since neither the final answer, nor the original question involves trig. Still...? How do we reconcile these gaps in continuity?

 

[Iruka]

That is an interesting question which I am still considering the answer to!

The integral is certainly defined, because it is simply the upper branch of the hyperbola y^2 - x^2 =4. Since it is a continuous curve, I can't see any reason why the area underneath would not be well defined. It is our substitution that screws (potentially, at any rate) things up.

I think it comes down to this: when you make the substitution, you are making certain assumptions about u (i.e, that u lies between 0 and pi/2), so that you can undo the substitution later.

But of course, u itself does not appear in the final answer.

I think the important thing to remember that as long as the function we get at the end of this process can be differentiated to yield the integrand, it doesn't actually matter how we got it in the first place.

 

[Iruka]

Upon further reflection, this problem with the trig substitution might be one reason why it is preferable to use a hyperbolic trig substitution.

I just asked maple to solve this question, and it gave the answer,

1/2*x*(4+x^2)^(1/2)+2*arcsinh(1/2*x)

so clearly maple prefers to use the hyperbolic trig substitution.

 

[Affinity]

tan substitutions work perfectly,
remember you don't get close to any funny points
since x = tan(u) ----> x varies over all reals, but tan(u) only has be defined for u in (-pi/2,pi/2) and for this interval secant is also fine.


there might potentially be problems with sec substitutions

 

[Slidey]

Ah, so you're saying that the substitution essentially squishes the domain from -infinity to infinity to (-pi/2,pi/2), and this is possible because both sets have cardinality aleph-null? And then we unsquish it at the end using arc functions...?

Well then, why wouldn't it work for sec? Sec is also analytic on the domain (-pi/2,pi/2), is it not?

Are these domains arbitrary? Could we just as easily say we're using the domain (-3pi/2,-pi/2)?

 

[Affinity]

x = sec(u) is fine when x is no less than 1 (or no greater than minus 1), else there might be problems...
for example int [-1 to 1] 1/x^2 does not exist but if you substitute x=sec(u) you get

int [pi to 0] sin(x) = -cos(0) + cos(pi) = -2


yeah you can use the domain -3pi/2 and -pi/2, you just need to define your inverse trig functions accordingly

 

[ngogiathuan]

And then we unsquish it at the end using arc functions...?

Well then, why wouldn't it work for sec? Sec is also analytic on the domain (-pi/2,pi/2), is it not?

We sub x = atan u in, and in the end we sub tan u = (x/a) and sec = sqrt (x^2+a^2) /a in to get a function in term of x. We dont use arc function
the substitution x = asec u doesnt simplify the integral so i dont think that we can use it. Maybe its possible but i havent tried yet

since x = tan(u) ----> x varies over all reals, but tan(u) only has be defined for u in (-pi/2,pi/2) and for this interval secant is also fine.

So when we let x =a tan (u) we restrict u from -pi/2 to pi/2. This satisfies that x varies over the real field and yet at the same time yelds sec u >0
So that solves my problem
yay thanks alot everyone
oh btw, the general formula for Integral of sqrt (x^2+a^2) (which i just looked up) is (x/2)*sqrt (P) + (a^2/2)ln(x+sqrt (P)) +C, if we let P = x^2 + a^2.
That book solved it using hyperbolic function

 

[Affinity]

you will have to use x=a*sec(x) now and then. for terms involving x^2 - 1
real numbers would probably be prefered to field, since here the algebraic properties are not of concern