Integration of a composite square root function, easy (hard for me )

[BlackWyvern]

Integration of a composite square root function, easy (hard for me :( )

Homework Statement

How would I integrate:

y = \sqrt{(1 + 4x^2)}

Homework Equations

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The Attempt at a Solution

I have no idea how to do this. I tried to to the chain rule in reverse, and then I ignored it, but I can't get the right answer. I don't properly understand integration. Just some nice working would be great.

 

[Gib Z]

\int \sqrt{ (2x)^2 +1} dx

Now, we know that \tan^2 x + 1 = \sec^2 x for all values of x. So the square root would disappear if we we're to make that 2x into tan u.

IE Use the Following substitution: x= (tan u)/2.

 

[BlackWyvern]

Is there a simpler way?

I know that for the special case where:

g'.f'(g) --> f(g)

Or something like that. The special case where the co-efficient is a multiple of the derivative of the inner function. But we don't have that here. That's fine, but what would be formula be for this case? All I see is people saying to use the chain rule in reverse. Well, I don't know how to do that. It's not as easy as putting the car into reverse and pressing accelerate.

 

[Gib Z]

Well the simpler way would be to use a general formula. Look here: http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions

\int \sqrt{x^2+a^2} \;dx = \frac{1}{2}\left(x \sqrt{x^2+a^2} +a^2\,\ln\left(x+\sqrt{x^2+a^2}\right)\right)

To get it to look like how it does in that formula, we have to take a factor of 4 from inside the square root, which is then a factor of 2 outside, and then take the constant of 2 out of the integral, to look like that:

2\int \sqrt{x^2 + (1/2)^2} dx.

Now just plug a=1/2 into the formula.

 

[BlackWyvern]

Okay, I'll start off with the entire question.

There is a parabola, y = x², and I want to find it's length from x= 0 to 10.

I was using the formula S = \int \sqrt{1 + (dy/dx)^2 } dx

So I get my dy/dx as 2x.

S = \int \sqrt{1 + (2x)^2 } dx

From my logic, I think that S is:

\sqrt {10^2 + (100)^2} < S < 10 + 100 == 100.499 < S < 110

The lower limit is the hypotenuse of a triangle from the origin to the point (10, 100). The upper limit is the length of both sides of the triangle. Is this logic incorrect?

 

[Kurdt]

Its just a bit of a complicated integral but it can be done. I wouldn't substitute tan, this is more easily tackled by making a sinh substitution. The answer will then be of the form Gib Z gave above. If its not important to show your working of the integral for this question then just look it up in a list f standard integrals as suggested above. If it is important try and work through slowly and find a web page with hyperbolic identities on it.

 

[BlackWyvern]

Thanks for the help. But I still can't seem to solve the original problem. It obviously seems that this is far out of my league, and it's nothing I'll be able to learn without understanding the simpler aspects of integration.

I'd like to be able to finish the problem for closure, however. :)

 

[BlackWyvern]

I get 100.30513 when I use that formula, outside of the limits I have. So I've done something wrong somewhere.

 

[BlackWyvern]

I used an applet with an approximation formula and 10,000 intervals, and I got an answer of 101.0473, which is within the limits geometry describes.

I'm not sure at all right now, I'll need to number crunch to see what does work.

Thanks very much for the help so far, I appreciate any further help.

 

[Kurdt]

When I use the given answer I get greater than the minimum estimate. I think perhaps you made a slight error when calculating.

 

[drpizza]

Since you're using an applet with an approximation formulas, may I suggest grabbing a graphing calculator and having it find the area under the curve f(x) = \sqrt{1 + (2x)^2 } from 0 to 10

TI 83+ gives 101.0473

 

[Kurdt]

I was going to suggest the OP ploughs through the integral using the sinh(u) = 2x substitution. I think it would help them a lot in understanding where these things come from especially if they're doing a course on vectors and arc lengths etc. So this is a plea to post some working and we can guide you through.

It can be kind of strange looking these standard integrals up and wondering where one obtains an arcsinh or logarithms from.

 

[bob1182006]

This is a popular integral for engineering etc...

You were going the right way through integration by parts:

u=sqrt(1+4x^2)
dv=1

You will get a new integral something like x^2/sqrt(1+4x^2) which will be split into 2 integrals, 1 being the original integral + or - a simple integral. so you'll have 2x Original Integral = something then you divide by 2 to get the answer.

Well if you already have your answer I can post the steps doing Integration by Parts.

 

[BlackWyvern]

I'm surprised that this was moved into the Calculus and Beyond section, I really thought I was dealing with the tip of the iceberg here.

I've been doing a small bit of thinking, and I'd just like to know, is it possible to use the substitution rule on the integral in its current form? Because if it is possible, then I will find out how. Unfortunately I can't find my Graphics calculator, so I can't see if I'm doing well or not. Even more unfortunately, I have some small exams on Friday. Thanks very much for the help so far, I haven't and won't give up on trying to understand this.

 

[bob1182006]

you could...but you'd haveta do 1+4x^2 = u, du=...dx solve for x to plug back into that equation which would be like..(u-1)/4 and then square root alll of that which would give you something like partial fractions I think.

Wouldn't be all that bad I guess. But if you do it by integration by parts you will surely get the answer.

 

[BlackWyvern]

I get this when I use substitution, is this correct? Or even on the right track?

\frac{2u^{3/2}} {3} x \frac{du}{8 \sqrt{\frac{u - 1}{4}}}

If it's wrong, please correct me. Also, how do I find the new limits of integration? I know I have to substitute the x values into somewhere, is it the u = f(x) that I substitute into? I think it is.

 

[bob1182006]

yea if you have 2 limits for x and you do a substitution u=f(x) then just plug in the limits into u=f(x) and you will get 2 new limits for u.

But wow I don't see a way to do the rest of the integration unless you do some trig substitution o.o but I'd love to see it be done!

 

[BlackWyvern]

\frac{2u^{3/2}} {3} x \frac{du}{8 \sqrt{\frac{u - 1}{4}}}

-->

\frac{u^{3/2}} {12 \sqrt{\frac{u - 1}{4}}} x du

Then, I am not sure. I thought I could simply substitute (1 + 4x²) for u.

 

[bob1182006]

u=1+4x^2? thatd make you go back where you were before if you used the substitution u=1+4x^2 before. If not then maybe another substitution might work, maybe...

Integration by Parts Will work btw I know it will.

 

[BlackWyvern]

I don't think I can use integration by parts as there is no product.

 

[BlackWyvern]

I used the integral that Gib Z gave to me earlier, but this time I did it well, and I received the correct answer of 101.047.

Now, I need to figure out how to derive this integral. I never stop until I find a method I can do in an exam if I forget the formula.

It was listed in the "List of irrational functions". I don't know what that means exactly, but I'm assuming it means they're impossible to integrate in a standard method, without some kind of infinite series.

 

[bob1182006]

No, irrational just means it can't be expressed as a fraction, ex \sqrt{2}

Well since you already have the answer (I hope). Here's how to do it through Integration by parts:

I=\int\sqrt{1+4x^2} dx
u=\sqrt{1+4x^2};v'=1

u'=\frac{4x}{\sqrt{1+4x^2}};v=x

I=x\sqrt{1+4x^2}-\int\frac{1+ 4x^2 dx -1}{\sqrt{1+4x^2}} added/subtracted 1 on the integral in order to separate it.

I=x\sqrt{1+4x^2}-\int\sqrt{1+4x^2} dx + \int\frac{dx}{\sqrt{1+4x^2}} Note: the original integral I is present here. Moving it to the left hand side.

2I=x\sqrt{1+4x^2}+\int\frac{dx}{\sqrt{\sqrt{1}^2+(2x)^2}}

\int\frac{dx}{\sqrt{x^2\pm a^2}}=ln|x+\sqrt{x^2\pm a^2}|

2I=x\sqrt{1+4x^2}+\int\frac{dx}{\sqrt{(2x)^2+\sqrt{1}^2}} Switched the numbers in the square root to get it in the proper form.

2I=x\sqrt{1+4x^2}+ln|(2x)+\sqrt{(2x)^2+\sqrt{1}^2}| Now to solve for I.

I=\frac{1}{2}x\sqrt{1+4x^2}+\frac{1}{2}ln|2x+\sqrt{4x^2+1}|

And that's how you do it. You can of course change 1 by a^2 in order to get the general form that Gib Z showed you.

 

[Kurdt]

I must say i still prefer the hyperbolic sine method. I will not post it here as I wanted the original poster to have a go. It is useful to know more than one method however.