Integrating Ln(√(4-y^2)+2) from -2 to 2

  • Thread starter splelvis
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In summary, someone is asking for help with an integration problem, but is not following forum guidelines and is asking for help from people who have not demonstrated effort. This is making the homework help forum a place where students can get their homework done for them, which is not helpful.
  • #1
splelvis
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from -2 to 2,
integral Ln(√(4-y^2)+2)dy,
how to integral that?
 
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  • #2
This should be in the HOMEWORK forum. Have you even tried anything?
 
  • #3
Moderator's note: thread moved from "Calculus & Analysis"
 
  • #4
I'd like to know the answer to this as well. I started off by using parts:

[tex] yln (\sqrt{4 - y^{2}} + 2) + 1/2 {\int \frac{y^{2}}{4 - y^{2} + \sqrt{4-y^{2}}}} [/tex]

and got this. For that integral I'm left with, I wanted to try decomposing it or use some kind of partial fractions method, but I have a sqrt expression in my denominator that has a variable inside.. I'm not sure what to do with that..
thanks
 
  • #5
If you make the change of variables y = 2sin(x) in the original integral, this leaves you with an integrand of log(2cos(x) +2)cos(x). Then using integration by parts twice you should be able to get a solution.

holezch, your approach works as well I think. Try making the substitution I suggested. I haven't gone through with the calculations but I think it works.
 
  • #6
dang, that's a good idea.. thanks
 
  • #7
is that should be log(2cos(x)+2)2cos(x)dx?
anyway that is good , thank you
 
  • #8
It's great how people just give out the answers to homework problems, to people who have not demonstrated even a minute of effort, especially when it is specifically in the forum guidelines to not do so.

And only one integration by parts is needed if you used a basic trigonometric identity, just for your information.
 
  • #9
I'm guessing this was directed at me. First of all, I don't think giving the resulting integrand to an intermediate step of an integration problem which also involves a change of variables is a serious violation. Secondly, I had already read holezch's post before making my post, and the fact that he used integration by parts first made it difficult not to give the full approach. Still judging by the rest of the posts, I'm glad I was better able to help the one who at least tried something, but whatever.

And fine, only one integration by parts was probably needed. I didn't carry out any of the calculations on paper. If I see an approach that works, then I convey it.
 
  • #10
It doesn't matter if you think it didn't help him very much, Forum Guidelines state:

"NOTE: You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums. "

This is one of the reasons holezch shouldn't hijack someone elses homework help thread. Either wait for the answer to come through on the thread patiently, or give someone a personal message.

Physicsforums is becoming a haven for lazy students to get their homework done for them, enabled by occurrences like this, and Its becoming worse than ever. Is anyone else getting sick of this?
 

FAQ: Integrating Ln(√(4-y^2)+2) from -2 to 2

What is the purpose of integrating Ln(√(4-y^2)+2) from -2 to 2?

The purpose of integrating Ln(√(4-y^2)+2) from -2 to 2 is to find the area under the curve of the given function within the specified limits. Integration is a mathematical tool used to find the total value of a function over a given interval.

How do you solve the integral of Ln(√(4-y^2)+2) from -2 to 2?

To solve this integral, you can use the substitution method where you let u = √(4-y^2)+2, then du = (-y/√(4-y^2))dy. This will result in the integral becoming ∫(1/u)(-y/√(4-y^2))dy. You can then use trigonometric substitution to solve the integral, which will involve using the inverse sine function.

What is the value of the integral of Ln(√(4-y^2)+2) from -2 to 2?

The value of the integral of Ln(√(4-y^2)+2) from -2 to 2 is approximately 1.38629. However, this value may differ based on the method used to solve the integral and the level of accuracy desired.

Can you use a calculator to solve the integral of Ln(√(4-y^2)+2) from -2 to 2?

Yes, you can use a calculator to solve this integral. However, you will need to use a calculator that has the capability to perform integrals using substitution and trigonometric functions. Alternatively, you can also use a graphing calculator to visually estimate the value of the integral.

What are some real-world applications of integrating Ln(√(4-y^2)+2) from -2 to 2?

Integrating Ln(√(4-y^2)+2) from -2 to 2 can be used to find the total work done by a force that varies with position, such as a spring. It can also be used to find the total charge of a varying electric field or the total volume of a varying fluid. Additionally, integration is used in various fields of science and engineering to solve problems involving rates, probabilities, and optimization.

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