# Evaluate the integral H.W

1. Aug 19, 2008

### afcwestwarrior

1. The problem statement, all variables and given/known data
∫arctan 4t dt

2. Relevant 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

2. Aug 19, 2008

### rock.freak667

$$\int arctan(4t) dt$$

u=arctan(4t)
dv=dt

not what you have.

3. Aug 19, 2008

### afcwestwarrior

du=??????
dv=t

4. Aug 19, 2008

### Feldoh

Whats the derivative of arctan(x)? Use the chain rule for arctan(4t)

5. Aug 20, 2008

### afcwestwarrior

the derivative for arctan x is 1/1+x^2 and the chain makes it look like (4) 4t/1+x^2

6. Aug 20, 2008

### rock.freak667

if x=4t

$$\frac{d}{dx}(arctan(x))=\frac{1}{1+x^2}$$

so get the differential of arctan(4t), you'll need to multiply $\frac{d}{dx}(arctan(x))$ by $\frac{dx}{dt}$. What does this give you?

7. Aug 20, 2008

### afcwestwarrior

it'll be 4t/1+x^2

8. Aug 20, 2008

### afcwestwarrior

man im confused

9. Aug 20, 2008

### Defennder

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. Aug 20, 2008

### rock.freak667

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. Aug 21, 2008

### Gib Z

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 thats not what he asked for.

12. Aug 21, 2008

### weejee

Hint :
u=arctan(4t), dv=dt

It'll work. (Just as you integrate log(x)).

13. Aug 21, 2008

### HallsofIvy

Staff Emeritus
Surely you understand that "arctan 4t" does NOT mean "arctan" time "4t"! "arctan" without a variable is meaningless.