Calculus Help

[zoooom]

Originally posted by: JohnCU

Originally posted by: zoooom

Originally posted by: JohnCU
integral table

I don't see any that would work with this particular problem.

it's one of the integrals of sqrt(u^2+a^2) type with u = x^(3/2) and a = 1 just need to find out how to work in the other element

With that substitution you end up getting (2/3)integ: du/sqrt(a^2+u^2) and we haven't learned a way to integrate that.

Edit: I forgot to put the 2/3 that the integral is multiplied by.

 

[chuckywang]

This is a horrible integral. You have to do it by parts and also factor 1+x^3 = (1+x)(1-x+x^2)

 

[JohnCU]

I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

 

[ruffilb]

Originally posted by: ruffilb
Oh man, if this is what I think it is, I LOVE these sorts of problems. It's complicated, but it's really cool.

I think it's a recursive integration by parts problem, where you have to pull over the int(Vdu) term to the other side and divide... can anyone confirm?

It's not, I screwed something up.

This is definitely not something you can do in your class. It's probably a typo 😉

 

[ruffilb]

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Eh? How does U then relate back to the original problem?

 

[JohnCU]

Originally posted by: ruffilb

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Eh? How does U then relate back to the original problem?

it's an indefinite integral so you substitute back whatever you substituted in in the first place.

 

[zoooom]

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Checking this now. =)

 

[chuckywang]

Originally posted by: ruffilb
the answer, btw, is

(2ln((x^3+1)^(1/2)+ x^(3/2))/3)

That's not the answer.

EDIT: Whoops, misread it...yeah that does work.

 

[zoooom]

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Hmm, yeah it does seem to work, but you forgot the +C. 🙂 I wasn't thinking the answer would be a hyperbolic.

 

[JohnCU]

Originally posted by: zoooom

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Hmm, yeah it does seem to work, but you forgot the +C. 🙂 I wasn't thinking the answer would be a hyperbolic.

lol, when you get to real practical uses of calculus no one uses the +C but yeah you are correct.

 

[zoooom]

Originally posted by: JohnCU

Originally posted by: zoooom

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Hmm, yeah it does seem to work, but you forgot the +C. 🙂 I wasn't thinking the answer would be a hyperbolic.

lol, when you get to real practical uses of calculus no one uses the +C but yeah you are correct.

Wtf, math has applications? :disgust: Seriously though, thank you to everyone to that tried to solve the integral, especially to JohnCU.

 

[chuckywang]

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Weird...my TI-89 says the integral of sqrt(1/(1+u^2)) is ln(sqrt(1+u^2)+u). But the integral is also arcsinh(u). I guess those two must differ by a constant.

 

[JohnCU]

Originally posted by: chuckywang

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Weird...my TI-89 says the integral of sqrt(1/(1+u^2)) is ln(sqrt(1+u^2)+x). But the integral is also arcsinh(u). I guess those two must differ by a constant.

yeah or its just two different forms of the same thing. easier to write arcsinh than all of that natural log jazz. 🙂

 

[chuckywang]

Originally posted by: JohnCU

Originally posted by: chuckywang

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Weird...my TI-89 says the integral of sqrt(1/(1+u^2)) is ln(sqrt(1+u^2)+x). But the integral is also arcsinh(u). I guess those two must differ by a constant.

yeah or its just two different forms of the same thing. easier to write arcsinh than all of that natural log jazz. 🙂

I know that y = sinh(x) = (e^x-e^-x)/2, so I was mildly surprised that you can add some constants to that and solve for x. Now I'm just trying to figure out how to do that.

 

[JohnCU]

Originally posted by: chuckywang

Originally posted by: JohnCU

Originally posted by: chuckywang

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Weird...my TI-89 says the integral of sqrt(1/(1+u^2)) is ln(sqrt(1+u^2)+x). But the integral is also arcsinh(u). I guess those two must differ by a constant.

yeah or its just two different forms of the same thing. easier to write arcsinh than all of that natural log jazz. 🙂

I know that y = sinh(x) = (e^x+e^-x)/2, so I was mildly surprised that you can add some constants to that and solve for x. Now I'm just trying to figure out how to do that.

hmm... i'm not sure, i'm kind of sleepy and maple isn't working but LMK what you find.

 

[chuckywang]

Originally posted by: JohnCU

Originally posted by: chuckywang

Originally posted by: JohnCU

Originally posted by: chuckywang

Originally posted by: JohnCU
I got it, u-substitution.

Let u = x^3/2
du = 3/2*x^1/2dx or dx = 2/3*du/x^1/2

the integral becomes 2/3*INT(1/sqrt(1+u^2)) and becomes 2/3*inverse sinh (x^3/2)

Weird...my TI-89 says the integral of sqrt(1/(1+u^2)) is ln(sqrt(1+u^2)+x). But the integral is also arcsinh(u). I guess those two must differ by a constant.

yeah or its just two different forms of the same thing. easier to write arcsinh than all of that natural log jazz. 🙂

I know that y = sinh(x) = (e^x-e^-x)/2, so I was mildly surprised that you can add some constants to that and solve for x. Now I'm just trying to figure out how to do that.

hmm... i'm not sure, i'm kind of sleepy and maple isn't working but LMK what you find.

Yep it works out. Let y = ln(sqrt(1+x^2)+x) and compute (e^y-e^-y)/2. It works out to equal x. BTW, I made a mistake: sinh(x) = (e^x-e^-x)/2.

You can also solve it directly like so:
If y=(e^x - e^-x)/2, then sqrt(1+y^2) = (e^x + e^-x)/2. Adding those together gives you e^x, so x = ln(y+sqrt(1+y^2)). Pretty cool trick having to do with exponentials and logarithms.

How the heck did you remember the derivative of arcsinh(x). Geez, I wasn't even taught the derivatives of arcsinh(x) and arccosh(x) when I took Calculus. All I was taught was that cosh(x) = (e^x+e^-x)/2 and sinh(x) = (e^x - e^-x)/2.