Is This the Correct Way to Integrate arctan(4t) by Parts?

  • Thread starter Thread starter afcwestwarrior
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The discussion revolves around the integration of the function arctan(4t) with respect to t, specifically using integration by parts. Participants are exploring the correct setup for applying this method.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to define the variables u and dv for integration by parts, with some suggesting different assignments. There is confusion regarding the differentiation of arctan(4t) and the application of the chain rule. Questions about the correctness of these assignments and the derivative of arctan are raised.

Discussion Status

There is an active exploration of different approaches to finding the integral, with some participants providing hints and others expressing confusion about the differentiation process versus integration. Multiple interpretations of the problem setup are being discussed.

Contextual Notes

Some participants note the importance of correctly understanding the function arctan(4t) and clarify that it should not be misinterpreted as a product of arctan and 4t. There is also mention of the need to express everything in terms of t.

afcwestwarrior
Messages
453
Reaction score
0

Homework Statement


∫arctan 4t dt


Homework Equations


integration by parts
∫u dv= uv- ∫v du


3. The attempt at a
u=arctan dv=4tdt
du=1/x^2+1 v=t/2

are these correct, the arctan throws me off
 
Physics news on Phys.org
[tex]\int arctan(4t) dt[/tex]

u=arctan(4t)
dv=dt

not what you have.
 
du=?
dv=t
 
Whats the derivative of arctan(x)? Use the chain rule for arctan(4t)
 
the derivative for arctan x is 1/1+x^2 and the chain makes it look like (4) 4t/1+x^2
 
afcwestwarrior said:
the derivative for arctan x is 1/1+x^2 and the chain makes it look like (4) 4t/1+x^2

if x=4t

[tex]\frac{d}{dx}(arctan(x))=\frac{1}{1+x^2}[/tex]

so get the differential of arctan(4t), you'll need to multiply [itex]\frac{d}{dx}(arctan(x))[/itex] by [itex]\frac{dx}{dt}[/itex]. What does this give you?
 
it'll be 4t/1+x^2
 
man I am confused
 
You're working in t, not x. As said earlier, use the chain rule for this. Let u=4t. d/dt arctan (4t) = du/dt d/du arctan(u). Express everything in t when you're done.
 
  • #10
Let y=arctan(4t).

Let u=4t.

So we have y=arctan(u). What is dy/du?What is du/dt?

EDIT: ahh Defennder, you type fast
 
  • #11
I'm pretty damn confused here as well. Half these posts are showing him how to find the derivative when he wants to know the antiderivative lol. I mean sure, its probably more important to know the derivative first, but that's not what he asked for.
 
  • #12
Hint :
u=arctan(4t), dv=dt

It'll work. (Just as you integrate log(x)).
 
  • #13
afcwestwarrior said:

Homework Statement


∫arctan 4t dt


Homework Equations


integration by parts
∫u dv= uv- ∫v du


3. The attempt at a
u=arctan dv=4tdt
du=1/x^2+1 v=t/2

are these correct, the arctan throws me off
Surely you understand that "arctan 4t" does NOT mean "arctan" time "4t"! "arctan" without a variable is meaningless.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
10
Views
2K