Integration help

  • Thread starter James889
  • Start date
  • #1
192
1
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.
 

Answers and Replies

  • #2
123
1
how about letting [tex]u=1+t^{2}[/tex]?
 
  • #3
192
1
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]
 
  • #4
123
1
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.
 
  • #5
192
1
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?
 
  • #6
123
1
Well, I think you are done. Don't you see it yet? Just expand and integrate.
 
  • #7
123
1
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:
  • #8
192
1
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]
 
  • #9
123
1
Don't forget 1/3
 

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