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

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Homework Help Overview

The problem involves evaluating the integral \(\int_0^2 \frac{4dy}{1+9y^2}\). The original poster expresses difficulty with integration techniques and mentions a tutor's suggestion regarding the presence of a tangent function in the solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss integration by substitution and the relevance of the derivative of arctan. There are attempts to rewrite the integral in a more familiar form, and questions arise about the correctness of the expressions derived during the discussion.

Discussion Status

The discussion is ongoing with various interpretations and attempts to clarify the integration process. Some participants provide insights into the derivative of arctan and its relation to the integral, while others express confusion about their results. There is no explicit consensus on the final answer, but guidance is being offered regarding the integration technique.

Contextual Notes

Participants are navigating through the integration process with some uncertainty about the application of limits and the correctness of their derived expressions. Miscommunication regarding the evaluation of limits has also been noted.

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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]
 
Last edited:
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?
 
  • #10
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?
 
  • #11
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]
 
  • #12
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?
 
  • #13
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.
 
  • #14
no problem, thank you very much for your assistance :)
 

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