Integral Help: t^3 &sqrt;(1+t^2)

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Homework Help Overview

The discussion revolves around the integral of the function t^3 multiplied by the square root of (1+t^2), specifically the expression ∫ t^3·√(1+t^2) dt. Participants express uncertainty regarding how to approach the integral due to the presence of the t^3 term.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the substitution u = 1 + t^2 as a potential approach. There is discussion about the need to express dt in terms of du and t^3 in terms of u. Some participants question the effectiveness of this substitution and express confusion about the next steps in the integration process.

Discussion Status

The conversation is ongoing, with participants providing guidance on how to proceed with the integral after the substitution. There are indications of misunderstanding regarding the transformation of variables, and some participants are correcting each other’s mistakes while attempting to clarify the steps needed to simplify the integral.

Contextual Notes

Participants note the necessity of having only one variable before integrating and the importance of correctly transforming dt and t^3. There is a recognition of potential mistakes in the application of the substitution, particularly regarding the coefficients involved.

James889
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Hi,

I would like some help with the following integral

[tex]\int t^3\cdot \sqrt{1+t^2}~dt[/tex]

I don't know what to do with the [tex]t^3[/tex] term.
If it wasn't there it would be easy.

Help appreciated.
 
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how about letting [tex]u=1+t^{2}[/tex]?
 
benhou said:
how about letting [tex]u=1+t^{2}[/tex]?

Thats what i thought about too, at first.

But it doesn't get me anywhere

[tex]t^3\cdot\sqrt{u}[/tex]
 
Well, before taking the integral, you have to have only one variable. What you need to do is to represent dt in terms of du, and t^3 in terms of u.
 
benhou said:
Well, before taking the integral, you have to have only one variable. What you need to do is to represent dt in terms of du, and t^3 in terms of u.

Aha!
I think i understand.

So i need [tex]\int(u-1)\sqrt{u}du[/tex]

How do i proceed now?
 
Well, I think you are done. Don't you see it yet? Just expand and integrate.
 
You made a mistake. since
[tex]u=1+t^{2}[/tex]
[tex]du=2tdt[/tex] or
[tex]dt=du/(2t)[/tex]

You should have 1/2 somewhere.
 
Last edited:
benhou said:
You made a mistake. since
[tex]u=1+t^{2}[/tex]
[tex]du=2tdt[/tex] or
[tex]dt=du/(2t)[/tex]

You should have 1/2 somewhere.

Oh, dang i forgot about that. Thanks!

[tex]\frac{1}{5}u^{5/2} -u^{3/2}[/tex]
 
Don't forget 1/3
 

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