Integration Using the arctan rule

[Wreak_Peace]

Homework Statement

Our question is \int (from 0 to 4) dt/ ((4t^2) + 9)
(For a better representation, because I fail at LaTeX: http://www.wolframalpha.com/input/?...&f5=4&f=DefiniteIntegralCalculator.rangeend_4 )

Homework Equations

So we have the rule:

\int du/(a^2 + u^2) = (1/a) arctan (u/a) + C

The Attempt at a Solution

So, using a = 3, and u = 2t (because the rule says that at the bottom, there is a^2 + u^2, so we take the square root of 9, and 4t^2)
I get 1/3 arctan( 2t/3) from 4 to 0
1/3 arctan( 8/3) - 1/3 arctan (0)
=.404..
However, Wolfram Alpha, and my graphing calculator seem to both give me .202, and wolfram alpha says the coefficient for the final solution is 1/6 arctan..., whereas I got 1/3 arctan..., and I cannot see how it becomes 1/6..

Thanks for reading!

 

[Dick]

Homework Statement

Our question is \int (from 0 to 4) dt/ ((4t^2) + 9)
(For a better representation, because I fail at LaTeX: http://www.wolframalpha.com/input/?...&f5=4&f=DefiniteIntegralCalculator.rangeend_4 )

Homework Equations

So we have the rule:

\int du/(a^2 + u^2) = (1/a) arctan (u/a) + C

The Attempt at a Solution

So, using a = 3, and u = 2t (because the rule says that at the bottom, there is a^2 + u^2, so we take the square root of 9, and 4t^2)
I get 1/3 arctan( 2t/3) from 4 to 0
1/3 arctan( 8/3) - 1/3 arctan (0)
=.404..
However, Wolfram Alpha, and my graphing calculator seem to both give me .202, and wolfram alpha says the coefficient for the final solution is 1/6 arctan..., whereas I got 1/3 arctan..., and I cannot see how it becomes 1/6..

Thanks for reading!

When you substitute u=2t, you also have to substitute du=2*dt. So also replace dt with du/2. That's the factor of two you are missing.

 

[Wreak_Peace]

Thanks, me and a couple of friends were stumped on that one.

 

[haruspex]

\int_0^4 \frac{dt}{4t^2 + 9}
...
So, using a = 3, and u = 2t

When you put u = 2t, what do you put for dt?