Solving Integration Problem - Get Help Now!

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Homework Help Overview

The discussion revolves around an integration problem where the original poster is struggling to match their answer with the one provided in a solution manual. The problem involves the integration of a function related to sine and requires the application of integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the half-angle formula and integration by parts (IBP) as potential methods to approach the problem. There are suggestions to break the problem into smaller parts to simplify the integration process. Questions arise regarding the change of variables and how it affects the limits of integration.

Discussion Status

Participants are actively engaging with the problem, offering advice on different methods and discussing the implications of variable substitution. There is a recognition of the challenges involved in the integration process, and some participants express a willingness to explore alternative approaches.

Contextual Notes

There is mention of a specific answer from a solution manual, which may influence the expectations of the original poster. The discussion also highlights a common hesitation among students regarding the change of limits during substitution in integration.

leroyjenkens
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I'm having a problem getting the book answer for this problem. The book answer is:
[tex]a^{2}(\frac{1}{3}-\frac{1}{2(n\pi)^{2}})[/tex]

I did all my work on paper and it saves me about a month of time to post the picture of the paper instead of typing everything out.

Anyone see a problem I made on the integration?
Thanks
 

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leroyjenkens said:
I'm having a problem getting the book answer for this problem. The book answer is:
[tex]a^{2}(\frac{1}{3}-\frac{1}{2(n\pi)^{2}})[/tex]

I did all my work on paper and it saves me about a month of time to post the picture of the paper instead of typing everything out.

Anyone see a problem I made on the integration?
Thanks

Well, for one thing, the half-angle formula (or whatever you prefer to call it) is [tex]\sin^2\theta=\frac{1}{2}(1-\cos2\theta),[/tex] so you kinda got off to a bad start there.

Edit - Also, if I could offer a bit of advice ... maybe break this problem into two smaller, easier (less prone to errors) problems:

(1) Use IBP to get a formula for the antiderivative/indefinite integral ##\int u^2\sin^2u\ du##

(2) Use a "full-blown" ##u##-sub (i.e. change the limits) and the formula from (1) to solve the definite integral that you were given.
 
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gopher_p said:
Well, for one thing, the half-angle formula (or whatever you prefer to call it) is [tex]\sin^2\theta=\frac{1}{2}(1-\cos2\theta),[/tex] so you kinda got off to a bad start there.

Edit - Also, if I could offer a bit of advice ... maybe break this problem into two smaller, easier (less prone to errors) problems:

(1) Use IBP to get a formula for the antiderivative/indefinite integral ##\int u^2\sin^2u\ du##

(2) Use a "full-blown" ##u##-sub (i.e. change the limits) and the formula from (1) to solve the definite integral that you were given.

Oh yeah, didn't notice I had the identity wrong, thanks.

The solution manual does that method you said, but with y's instead of u's. I wanted to use the method I would have used on a test, but it's so long. I'd like to use the method the solution manual uses, but I'm not quite sure what they did. They made y the argument of the sine, and set the limits from 0 to n[itex]\pi[/itex] instead of a. How does that work? What's it called? I'd like to look it up. Thanks.
 
leroyjenkens said:
Oh yeah, didn't notice I had the identity wrong, thanks.

The solution manual does that method you said, but with y's instead of u's. I wanted to use the method I would have used on a test, but it's so long. I'd like to use the method the solution manual uses, but I'm not quite sure what they did. They made y the argument of the sine, and set the limits from 0 to n[itex]\pi[/itex] instead of a. How does that work? What's it called? I'd like to look it up. Thanks.

When you make the substitution/change of variables ##y=\frac{n\pi}{a}x##, then [tex](1)\ \frac{a}{n\pi}y=x\text{ and }\frac{a}{n\pi}dy=dx,[/tex] and [tex](2)\ y=0\text{ when }x=0\text{ and }y=n\pi\text{ when }x=a.[/tex]

Then [tex]\frac{2}{a}\int\limits_0^ax^2\sin^2(\frac{n\pi}{a}x)\ dx=\frac{2}{a}\int\limits_0^{n\pi}(\frac{a}{n\pi}y)^2\sin^2(y)\ \frac{a}{n\pi}dy[/tex]

This (changing the limits of integration) is the part of basic ##u##-sub that most students in my experience (myself included) are hesitant to learn because it's "easier" to just back-sub and use the old limits. In reality, it's not that hard once you realize what you need to do and buy into the fact that it's that easy. Also, learning to do it this way (and kinda understanding what is happening) will make some concepts in multivariate calc much simpler to grasp if that's something that is in your future.

I totally dig that you want to practice doing problems the way that you''ll need to do them on the test. I would still encourage you to just try it out. You might find that my way is faster. You might find that the trade off in time spent is worth it to minimize mistakes. In this case, I personally would at the very least make the substitution. Just be aware that the option of cutting it into two smaller problems is there, and it sometimes is the way to go.
 
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gopher_p said:
When you make the substitution/change of variables ##y=\frac{n\pi}{a}x##, then [tex](1)\ \frac{a}{n\pi}y=x\text{ and }\frac{a}{n\pi}dy=dx,[/tex] and [tex](2)\ y=0\text{ when }x=0\text{ and }y=n\pi\text{ when }x=a.[/tex]

Then [tex]\frac{2}{a}\int\limits_0^ax^2\sin^2(\frac{n\pi}{a}x)\ dx=\frac{2}{a}\int\limits_0^{n\pi}(\frac{a}{n\pi}y)^2\sin^2(y)\ \frac{a}{n\pi}dy[/tex]

This (changing the limits of integration) is the part of basic ##u##-sub that most students in my experience (myself included) are hesitant to learn because it's "easier" to just back-sub and use the old limits. In reality, it's not that hard once you realize what you need to do and buy into the fact that it's that easy. Also, learning to do it this way (and kinda understanding what is happening) will make some concepts in multivariate calc much simpler to grasp if that's something that is in your future.

I totally dig that you want to practice doing problems the way that you''ll need to do them on the test. I would still encourage you to just try it out. You might find that my way is faster. You might find that the trade off in time spent is worth it to minimize mistakes. In this case, I personally would at the very least make the substitution. Just be aware that the option of cutting it into two smaller problems is there, and it sometimes is the way to go.

Edit: I understand for now.

Thanks.
 

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