Integration Q

[Molondo]

Intergrate Sqrt(x^2 + 1 )



Internet doesn't have a clear answer, perhaps ATOT does?!

 

[TestedAcorn]

2.468

 

[BALIstik916]

think of Sqrt(x^2 + 1) as (x^2 + 1)^1/2
then use the power rule so...
2/3 (x^2 + 1)^3/2 and use the chain rule for the inside? not sure havent done in a while

 

[Molondo]

Originally posted by: BALIstik916
think of Sqrt(x^2 + 1) as (x^2 + 1)^1/2
then use the power rule so...
2/3 (x^2 + 1)^3/2 and use the chain rule for the inside? not sure havent done in a while

You are thinkin of Derivatives. I'm asking about antiderivitive.

 

[frostedflakes]

If you didn't know already, I think you have to use trig substitution to solve. Without looking it up though I wouldn't remember how to solve it. Been a while since I've done calc 2, now I just let computers do the integration.

 

[BALIstik916]

Originally posted by: Molondo

Originally posted by: BALIstik916
think of Sqrt(x^2 + 1) as (x^2 + 1)^1/2
then use the power rule so...
2/3 (x^2 + 1)^3/2 and use the chain rule for the inside? not sure havent done in a while

You are thinkin of Derivatives. I'm asking about antiderivitive.

umm the derivative of x^1/2 would be 1/2 x^-1/2
the antiderivative of x^1/2 would be 2/3 x^3/2 right?

 

[chuckywang]

Originally posted by: Molondo
Intergrate Sqrt(x^2 + 1 )



Internet doesn't have a clear answer, perhaps ATOT does?!

n/m, i made a mistake ... correcting it now.

 

[frostedflakes]

FWIW I plugged it into my TI-89 and it gives (ln(sqrt(x^2+1)+x))/2 + (x*sqrt(x^2+1))/2 + C

Like I mentioned I'm pretty sure you have to use trig substitution.

edit: Plugged something in wrong, fixed answer

 

[chuckywang]

Originally posted by: frostedflakes
FWIW I plugged it into my TI-89 and it gives (x^3)/3 + x + C

Like I mentioned I'm pretty sure you have to use trig substitution.

Dude, you forgot the square root.

 

[frostedflakes]

What, no I didn't

 

[chuckywang]

Originally posted by: frostedflakes
What, no I didn't

You did the integral of x^2 + 1, not the integral of sqrt(x^2 + 1).

 

[blinky8225]

I'm pretty sure you have to use Trig substitution and then follow that with integration by parts. The integral is giving me a little trouble, though.

 

[esun]

x = tan(theta)
dx = sec^2 (theta) dtheta
theta = arctan(x)

int(sqrt(1 + x^2) dx) = int(sqrt(1 + tan^2(theta)) sec^2(theta) dtheta)

= int(sec(theta) sec^2(theta) dtheta)
= int(sec^3(theta) dtheta)

I think that can be tackled with integration by parts, but I'm too lazy to finish it up (note that I used 1 + tan^2(theta) = sec^2(theta) above). This is based on this method:

http://en.wikipedia.org/wiki/Trigonometric_substitution

 

[Fenixgoon]

trig substitution is your friend

 

[eLiu]

Originally posted by: frostedflakes
FWIW I plugged it into my TI-89 and it gives (ln(sqrt(x^2+1)+x))/2 + (x*sqrt(x^2+1))/2 + C

Like I mentioned I'm pretty sure you have to use trig substitution.

edit: Plugged something in wrong, fixed answer

That's a pretty un-enlightening form of the result. I'm guessing that y'all haven't figured out that (ln(sqrt(x^2+1)+x)) = arcsinh(x) where arcsinh is the inverse hyperbolic sine; sinh = (exp(x) - exp(-x))/2

Anyway when you're dealing with 1/sqrt(a^2-x^2) type terms, you substitute x = sin(u) b/c d(arcsin(x))/dx = 1/sqrt(1-x^2)

When you have the "opposite" with sqrt(a^2+x^2) type terms, you substitute x = sinh(u) b/c d(arcsinh(x))/dx = sqrt(1+x^2)

Try it out and see. A few useful identities: cosh(x)^2 - sinh(x)^2 = 1. d(sinh(u))/du = cosh(u) and d(cosh(u))/du = sinh(u). So you can turn the integrand into int(cosh(u)^2 du) which should be quite simple.

-Eric

edit: what esun put also works, but it's ridiculously more complicated than using hyperbolic functions. In that method, you have to figure out the integral of sec(x)^3 (simplify with identities or use by-parts), and you'll have to evaluate terms like sec(arctan(x)) (hint: draw a triangle, label what all these trig things actually mean, and evalulate geometrically).

 

[JohnCU]

i can't believe how easily i have forgotten this stuff

 

[frostedflakes]

What are hyperbolic trig functions used for anyways? I've heard of them, but obviously not very familiar with them. I'm an engineer, not a math major, so cut me some slack.

 

[JohnCU]

Originally posted by: frostedflakes
What are hyperbolic trig functions used for anyways? I've heard of them, but obviously not very familiar with them. I'm an engineer, not a math major, so cut me some slack.

you should have covered this in calculus II

 

[frostedflakes]

I remember seeing hyperbolic functions, but that's about it. I don't think we went into any great detail on them, the instructor just defined some of the basic functions like sinh and cosh. I took it during the summer, maybe they skipped over some stuff. Might have also seen them in diff EQ, we did a bit of complex analysis in there (that stuff started to go way over my head, though).

 

[blinky8225]

With esun's method...

Int[Sqrt[x^2+1]dx]

Let x = tan[theta]
dx = sec^2[theta]d theta

Using tan^2[theta] + 1 = sec^2[theta]

You now have... Int[sec^3[theta]d theta]
Integrate by parts...
Let u = sec[theta] and dv = sec^2[theta]d theta

So you have...
Sec[theta]tan[theta] - Int[sec[theta]tan^2[theta] d theta]

sec[theta]tan^2[theta] = sin^2[theta]/cos^3[theta] = (1 - cos^2[theta])/cos^3[theta] =
Int[sec^3[theta] d theta] - Int[sec[theta] d theta]

You can do Int[sec[theta] d theta] by multiplying top and bottom by (sec[theta] + tan[theta]) and doing a change of variable where u = (sec[theta] + tan[theta]).

Move Int[sec^3[theta] d theta] to the other side divide by two and then substitute x back in.

Your answer will math the one given above by frosted flakes.

 

[eLiu]

Originally posted by: frostedflakes
What are hyperbolic trig functions used for anyways? I've heard of them, but obviously not very familiar with them. I'm an engineer, not a math major, so cut me some slack.

They come up uh... rarely, lol. Oftentimes when you use hyperbolics you could just as easily use the equivalent exponential. This is a lot less convenient with trig functions.

Regular/hyperbolic functions are really rather closely related. In a sense, hyperbolic trig functions are like "real" trig functions since cosh(ix) = cos(x):
sin(x) = (exp(ix) - exp(-ix))/2i, cos(x) = (exp(ix) + exp(-ix))/2
The exp(ix) terms (i means sqrt(-1) here) are what produce the oscillatory nature.
sinh(x) = (exp(x) - exp(-x))/2, cosh(x) = (exp(x) + exp(-x))/2
The hyperbolic trig functions are kinda like two exponentials back-to-back. For x >> 0 or x << 0, one of the exponential terms dominates while the other will be oppositely small.

Oh and the reason they're called hyperbolic trig functions is b/c (cosh(x),sinh(x)) traces out a hyperbola. Recall that (cos(x),sin(x)) traces out the unit circle.

I guess only times I've really seen them are:
1) integration tricks (like above)
2) solution of some ODEs: for example, the equation that describes how a rope will hang. The shape is called a "catenary" and it's decribed by cosh. (Google for more info)
3) solutions of Laplace's equation which applies to like, potential flow/gravitational fields/electromagnetics and heat transfer (b/c laplace's eqn is like a steady-state heat transfer problem).

 

[frostedflakes]

Cool, thanks for taking the time to write up an explanation.

 

[Aharami]

i used to find calculus so easy. now it's all gibberish to me

 

[eLiu]

Originally posted by: frostedflakes
Cool, thanks for taking the time to write up an explanation.

Sure, np.

lol I just realized that my first post sounds kinda ass. I wasn't criticizing you, I was criticizing the TI-89 for not giving a more 'elegant' expression. (I mean, not that it really matters.)

 

[chuckywang]

You have to do trig substitution and integration by parts:

Let x = tan(?), so dx = sec^2(?)d?

Therefore, the integral becomes: sqrt(1+tan^2(?))*sec^2(?)d? = sec^3(?)d?.

Now integrate by parts. Let u' = sec^2(?) d?, so u = tan(?), and v = sec(?), so v' = sec(?)tan(?).
The integral becomes:
sec(?)*tan(?) - integral(sec(?)*tan^2(?)d?)
= sec(?)*tan(?) - integral((sec^3(?) - sec(?))d?).

We can move the integral of sec^3(?) to the other side:
So integral(sec^3(?) d?) = 1/2*(sec(?)tan(?) + integral(sec(?)d?).

I'm gonna assume if you're doing this problem, you already know that integral of secant is ln|sec(?) + tan(?)|.

Therefore, integral(sec^3(?) d?) = 1/2*(sec(?)tan(?) + ln|sec(?) + tan(?)|.

Substitute for x, and you're done!!