Solving Difficult Integral - Tom's Struggles

[Kb1jij]

I have been stumped trying to find the following integral:

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

I put it into my TI-89 calculator and it gave me and answer (that checks), but I cannot figure out how to do it by hand (is it possible?).

I tried using substitution and got:

\int \frac{\sqrt{u}}{\sqrt{u-1}} du

If you then use integration by parts, it just flips the fraction. Any suggestions?

Thanks!
Tom

 

[Hootenanny]

Does writing it as

\int (1+x^{2})^{\frac{1}{2}}

Help?

 

[robphy]

What substitution did you try?

 

[Hootenanny]

What substitution did you try?

I think he used u = 1+x^2 becasue I've just tried it and arrived at the same result.

 

[Kb1jij]

You can also try starting with integration by parts, and that gives you:

x \sqrt{1+x^2} - \int \frac{x^2}{\sqrt{1+x^2}} dx

but that doesn't help

 

[Hootenanny]

For this type of problem you can use the inverse of the chain rule to give;

\int (ax + b)^n \;\; dx = \frac{1}{a(n+1)}\cdot (ax+b)^{n+1} +c

Hope this helps
-Hoot

 

[TD]

Have you seen the hyperbolic functions? You can use cosh²a-sinh²a = 1 to substitute x = sinh(a).
If you haven't seen those, using 1+tan²a = sec²a is an alternative to use x = tan(a).

 

[nrqed]

For this type of problem you can use the inverse of the chain rule to give;

\int (ax + b)^n \;\; dx = \frac{1}{a(n+1)}\cdot (ax+b)^{n+1} +c

Hope this helps
-Hoot

But he (she) has a x^2 so this does not help... (or I might be missing something). A trig substitution seems the way to go to me...

Pat

 

[Hootenanny]

But he (she) has a x^2 so this does not help... (or I might be missing something). A trig substitution seems the way to go to me...

Pat

Okay I could also write it like this;

\int [f(x)]^n \;\; dx = \frac{1}{f'(x)\cdot (n+1)} \cdot [f(x)]^{n+1}

I trig ident is not required . I could show the full derivation if the OP likes.

 

[nrqed]

Okay I could also write it like this;

\int [f(x)]^n \;\; dx = \frac{1}{f'(x)\cdot (n+1)} \cdot [f(x)]^{n+1}

I trig ident is not required . I could show the full derivation if the OP likes.

I am sorry, but this is incorrect! (*unless f'(x) is a constant!)

Pat

 

[Hootenanny]

I am sorry, but this is incorrect! (*unless f'(x) is a constant!)

Pat

Indeed you are correct. I;ve obviuosly not been listening in class

 

[nrqed]

Have you seen the hyperbolic functions? You can use cosh²a-sinh²a = 1 to substitute x = sinh(a).
If you haven't seen those, using 1+tan²a = sec²a is an alternative to use x = tan(a).

To Kb1jij:

TD is right.
That's indeed the way to go. Using x= tan (\theta), one gets {\sqrt {1 + x^2}} = sec (\theta).

Then find dx in terms of d \theta. The resulting integral is easy to do. Then you need to reexpress \theta in terms of x.

Pat

 

[Kb1jij]

To Kb1jij:

TD is right.
That's indeed the way to go. Using x= tan (\theta), one gets {\sqrt {1 + x^2}} = sec (\theta).

Then find dx in terms of d \theta. The resulting integral is easy to do. Then you need to reexpress \theta in terms of x.

Pat

So the result is \int sec^3 (\theta) d \theta, right?
I don't think that this is really "easy to do", but I'll let you know if I get stuck.

Thanks!
Tom

 

[nrqed]

So the result is \int sec^3 (\theta) d \theta, right?
I don't think that this is really "easy to do", but I'll let you know if I get stuck.

Thanks!
Tom

You are right, it does not look pretty. Sorry.

Then, you must use hyperbolic trig functions. Set x = sinh(u). You will get the integral of cosh(u)^2. Using the hyperbolic trig identity for cosh^2, you will get a very easy integral to do.

Pat

 

[Kb1jij]

You are right, it does not look pretty. Sorry.

Then, you must use hyperbolic trig functions. Set x = sinh(u). You will get the integral of cosh(u)^2. Using the hyperbolic trig identity for cosh^2, you will get a very easy integral to do.

Pat

Ok, I found the integral, and I am left with this:
\frac{\theta + cosh \theta sinh \theta}{2}
which can then be turned into (arcsinh x + x cosh(arcsinh x))/2. Can I simplify that further? I'm not too familiar with the hyperbolic trig functions, but if those were regular trig functions that second term would become
x\sqrt{1-x^2}. Can I do the same here?

 

[Hurkyl]

\int \sec^3 \theta \, d\theta
is one of the standard (circular) trigonometric integrals. Your textbook probably explicitly states an algorithm for simplifying integrals of the type

\int \sec^a \theta \tan^b \theta \, d\theta

and the one of interest is certainly in any integral table.

I think the recommended step is to rewrite the integrand using \sec^2 \theta = 1 + \tan^2 \theta.


As for the hyperbolic substitution...

Can I do the same here?

You can use the same idea that was used to derive that result for the circular trig functions. However, the result will probably not be identical.

 

[nrqed]

Ok, I found the integral, and I am left with this:
\frac{\theta + cosh \theta sinh \theta}{2}
which can then be turned into (arcsinh x + x cosh(arcsinh x))/2. Can I simplify that further? I'm not too familiar with the hyperbolic trig functions, but if those were regular trig functions that second term would become
x\sqrt{1-x^2}. Can I do the same here?

Do not turn it back in terms of x yet! You must still integrate! *After* you have integrated you will go back to x. It's easy to integrate because cosh is the derivative of sinh! (or vice versa). That's the beauty of this trick...

 

[nrqed]

\int \sec^3 \theta \, d\theta
is one of the standard (circular) trigonometric integrals. Your textbook probably explicitly states an algorithm for simplifying integrals of the type

\int \sec^a \theta \tan^b \theta \, d\theta

and the one of interest is certainly in any integral table.

I think the recommended step is to rewrite the integrand using \sec^2 \theta = 1 + \tan^2 \theta.


As for the hyperbolic substitution...

You can use the same idea that was used to derive that result for the circular trig functions. However, the result will probably not be identical.

I don't think that this trick works if the difference of exponents a-b is 3. It works only for specific differences of exponents (I am too busy right now to work it out)... But I may be missing something in which case I would be interested in seeing how this works out...

Pat

 

[Hurkyl]

Oh right, I made a mistake. (That's what I get for doing it in my head) I needed to also use integration by parts to evaluate \int \sec^3 \theta \, d\theta.

 

[Curious3141]

If I'm not mistaken, we had a thread very similar to this one a while back. I came to the conclusion that hyperbolic trig was the most direct and elegant way. Let me search...

 

[Hurkyl]

Problems can vary. If I do one substitution and am not immediately happy with the results, I will often try the other substitution to see if I like it better. With some I prefer the hyperbolic substitution, and others have turned out nicer with a circular substitution.

 

[Curious3141]

OK, found it : https://www.physicsforums.com/showthread.php?t=104258&highlight=hyperbolic

Not exactly the same problem, but similar. VietDao showed a very insightful way to integrate that horrid secant and tangent term that allowed one to use conventional trig easily.

But the hyperbolic trig is still easier to "see", IMVHO.

An even cooler way would be to subsitute x = iu and transform it into a circular trig function using complex arguments (that's actually equivalent to the hyperbolic trig approach).

 

[VietDao29]

OK, found it : https://www.physicsforums.com/showthread.php?t=104258&highlight=hyperbolic

Not exactly the same problem, but similar. VietDao showed a very insightful way to integrate that horrid secant and tangent term that allowed one to use conventional trig easily.

Nah, it's not very insightful...
In case you don't want a hyperbolic substitution, you can try the following way. The first step is to find:
\int \frac{dx}{\sqrt{1 + x ^ 2}}
Let:
\sqrt{1 + x ^ 2} = t - x
Take the differential of both sides gives:
\frac{x}{\sqrt{1 + x ^ 2}} dx = dt - dx
\Rightarrow \frac{x + \sqrt{1 + x ^ 2}}{\sqrt{1 + x ^ 2}} dx = dt
\Rightarrow \frac{t}{\sqrt{1 + x ^ 2}} dx = dt
Divide both sides by t, then integrate both sides gives:
\Rightarrow \int \frac{dx}{\sqrt{1 + x ^ 2}} = \int \frac{dt}{t} = \ln |t| + C = \ln |x + \sqrt{1 + x ^ 2}| + C
Then let's find:
\int \sqrt{1 + x ^ 2} dx. By Integrating by Parts, we have:
\int \sqrt{1 + x ^ 2} dx = x \sqrt{1 + x ^ 2} - \int \frac{x ^ 2}{\sqrt{1 + x ^ 2}} dx = x \sqrt{1 + x ^ 2} - \int \frac{x ^ 2 + 1 - 1}{\sqrt{1 + x ^ 2}} dx
= x \sqrt{1 + x ^ 2} + \int \frac{dx}{\sqrt{1 + x ^ 2}} - \int \sqrt{1 + x ^ 2} dx.
Now, you can go from here, right? :)

 

[HallsofIvy]

Although I feel sure the hyperbolic substitution is easier, here is another standard way to integrate \int sec^3(\theta) d\theta:
\int sec^3(\theta)d\theta= \int \frac{d\theta}{cos^3(\theta)}= \int\frac{cos(\theta)d\theta}{cos^4(\theta)}
= \int \frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}
Let u= sin(\theta)[/tex] and the integral becomes<br /> \int \frac{du}{(1-u^2)^2}<br /> which can be done with partial fractions.

 

[Kb1jij]

Great, thanks guys! I understand both methods. I think a trig substitution is a little easier though. Either way I came out with a final answer of:
\frac{x \sqrt{1+x^2} + ln |x+\sqrt{1+x^2}|}{2} + C

Thanks again!
Tom