Integration Problem: Solve \int4dy/(1+9y^{2}) 2,0

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Homework Statement



Determine [tex]\int[/tex]4dy/(1+9y[tex]^{2}[/tex]) With limits of 2,0.

Homework Equations





The Attempt at a Solution



Have attempted ingtegration by substitution but have had no luck solving this problem. A maths tutor who went over it very quickly established there was a tan in the answer, i have not integrated anything like this before so don't really know where to start.
 
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do you know what the derivative of arctan is?
 
maybe this looks a little more familiar

[tex]\int_{0}^{2}\frac{4dy}{1+(3y)^{2}}[/tex]
 
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I have just looked up the definition, can't quite see how it will fit
 
[tex]4\int_{0}^{2}\frac{dy}{1+(3y)^{2}}[/tex]

[tex]\mbox{Let u=3y}[/tex]

does it look a little more familiar now?
 
I ended up with

4arctan(6)

Am i close?
 
no, example

[tex]\frac{d}{dy}\tan^{-1}(3y^{2})=\frac{6ydy}{1+(3y^{2})^{2}}[/tex]
 
Hmm i can't seem to get it, when i integrate i get

[tex]\frac{1}{12}tan^{-1}(12)[/tex]
 
Ignore that last post, is

[tex]\frac{4}{3}tan^{-1}(6)[/tex] correct?
 
do you notice the pattern with my problem?

the angle is [tex]3y^{2}[/tex]

where did my angle and derivative end up when i differentiated?
 
tunabeast said:
Ignore that last post, is

[tex]\frac{4}{3}tan^{-1}(6)[/tex] correct?
you're constants are correct but you're angle is wrong. if i took the derivative of your problem it would end up being 0 b/c you're basically saying it's a constant.

[tex]\frac{4}{3}\frac{0}{1+36}[/tex]
 
The 6 is just the value of the limits substituted into get a final answer, or is not that what the substituted value would be?
 
tunabeast said:
The 6 is just the value of the limits substituted into get a final answer, or is not that what the substituted value would be?
yes that is correct, i did not realize you were already plugging your limits in and evaluating. sorry, miscommunication.
 
no problem, thank you very much for your assistance :)