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Integration troubles

  1. Jan 24, 2014 #1
    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
     

    Attached Files:

  2. jcsd
  3. Jan 24, 2014 #2
    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.
     
    Last edited: Jan 24, 2014
  4. Jan 24, 2014 #3
    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.
     
  5. Jan 24, 2014 #4
    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.
     
  6. Jan 25, 2014 #5
    Edit: I understand for now.

    Thanks.
     

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