- #1

- 192

- 1

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.

- Thread starter James889
- Start date

- #1

- 192

- 1

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.

- #2

- 123

- 1

how about letting [tex]u=1+t^{2}[/tex]?

- #3

- 192

- 1

Thats what i thought about too, at first.how about letting [tex]u=1+t^{2}[/tex]?

But it doesn't get me anywhere

[tex]t^3\cdot\sqrt{u}[/tex]

- #4

- 123

- 1

- #5

- 192

- 1

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.

[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

Oh, dang i forgot about that. Thanks!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.

[tex]\frac{1}{5}u^{5/2} -u^{3/2}[/tex]

- #9

- 123

- 1

Don't forget 1/3

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