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Integration Q

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
 
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
Click to expand...

You are thinkin of Derivatives. I'm asking about antiderivitive.
 
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. 😛
 
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
Click to expand...

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

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?
 
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
 
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.
Click to expand...

Dude, you forgot the square root.
 
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.
 
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
 
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
Click to expand...

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).
 
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. 😛
 
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. 😛
Click to expand...

you should have covered this in calculus II
 
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).
 
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.
 
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. 😛
Click to expand...

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

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.)
 
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!!
 
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