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Question arrising from integration homework (advanced integration i guess?)

  1. Oct 24, 2004 #1
    Hi, so I've got a simple problem to evaluate the surface area for 1/x when rotated around the x axis. Ok, so that takes a quit bit of the magic and you wind up with the integral of (x^4+1)^1/2 over x^3. now with a substition you could make that into a form, sqrt(x2+1) over x2.

    That's what the book says to do. I'm a dork though and I don't want to use a 'form'. I mean I have to remember enough already, I'd rather just work though the problems. So I decide to tackle the bastard with some trigonometric substitution. I end up with sec^3 x over tan^2 x. I work that through into an integration by partial fractions problem, and blamo 1/1-u^2 + 1/u^2.

    Ok great on the u^-2, but now i've got another integration problem that I have to use a form for -- (1-u^2)^-1. So I tried to put sin x back into u. That winds up asking me to integrate sec x -- blam that one sucks. I've tried integration by parts and the third generation made we think I did something wrong. Anyone familiar with a proof of that and willing to give me any hints on the proper strategy? (on either the proof for (1-u^2)^-1 or sec x)

    So while writing this I got an idea that I could try substituting cos x and doing another substitution instead of going back one. I think i've done the homework problem for the integral of csc x. I'm going to do that but I have to get some housework done, go to work and study for physics exam.

    It's still going to bug me, I'd really be great for anyone and their ideas, below is the link to the work i've done in pdf if you want to see. The last line should be u^-2 it's a mistake.

    http://mypage.iu.edu/~nlcooper/math!.pdf [Broken]

    thanks so much for anyone who can read through my buzzed mathmatical excursions. is differential stuff as fun? i'm looking forward to those and debating on changing my major from physics to math.
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Oct 24, 2004 #2
    oh, and i dropped a pi at the last line, i forgot i even put it into parenthesis so i could add it in.
  4. Oct 24, 2004 #3
    wouldent that be the integral from -infinity to infinity of 1/Y dx..
    since you are revolving around the x axis...
  5. Oct 24, 2004 #4
    well the homework gaves a domain of 1 to 3 or something like that. i did the homework problem using the retarded forms, but want to do it the real way. in that case i simplified, making it an integral where the range can be specified after the entire integration process. i don't care about filling in stuff anyway, i'm not in algebra anymore. i just do it when i have to.
  6. Oct 24, 2004 #5

    Tom Mattson

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    No, the formula for the surface area is S=2π∫r(x)[1+(f'(x))2]dx.


    The integral is pretty easy if you do it by parts. Start from:


    and let u=(1+x4)1/2 and dv=dx/x3.
  7. Oct 24, 2004 #6
    i have the solution manual and this is the correct answer up to the part before integration by parts. maybe its around the y axis instead, but i don't think so

    my substituions were a different style than the solution manual, but at the point sqrt(x^4+1)/x^3 i chose to bypass subsituting u^2 for x^4, which would let me use a form sqrt(a^2+u^2)/u^2

    basically i guess i wanted to prove that since i didn't understand the way it works out. my book just 'verifies' it. so i start the integration and i beleive its correct to 1/(1-u^2)

    that integration hint you gave me is awesome, and i will attempt to work it out and learn a new style for something like that. thank you very much.

    but i was able to get very far with trig substitiution, i really like using it and i'd like to finish by finding the proof for (1-x^2) or sec x or hints on those differentiations. unless of course there's an error on my pdf (if anyone who has time would look at it you'd be my hero)

    i really like math and want to be able to do very well, grad school is a long way away.
  8. Oct 24, 2004 #7

    Tom Mattson

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    Trig substitution is a powerful tool, but it is just that: a tool. And as always, it is important to use the right tool for the right job. I tried to continue your work with the trig substitution, and it was unworkable as a secant-tangent type trig integral. I then converted it to a sine-cosine integral, and it was no better. Since those are the only 2 options I can see for continuing, I conclude that using trig substitution for this integral is as inappropriate as using a hacksaw to tighten a screw.

    But if you use integration by parts, the answer comes out almost immediately.
  9. Oct 24, 2004 #8

    Tom Mattson

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    Whoops, your partial fraction expansion does make this approach doable.

    However, it's still a lot more work than integrating by parts.
  10. Oct 24, 2004 #9
    yes i am going to work out that method tommorow after my physics exam. i'm primarily interested in integrating sec x. i am almost 100% i can do csc x because i am certain it was a homework problem from earlier in the semester.

    i don't memorize the formula things in the back of my calc book. i try i guess somewhat prove them, and then i know how to tackle that type of problem whenever it may arise. using integration by parts hadn't occured to me since it looked so ugly. the answer key and my professor used a similar technique to what i used. i will have to start working with that some more to see if i can apply it more often, i seem to prefer those crazy trig functions.

    thanks for the idea again
  11. Oct 25, 2004 #10

    Tom Mattson

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    If you can do csc(x), then you can do sec(x). The trick is nearly identical.

    In the case of sec(x), multiply the integrand by [sec(x)+tan(x)]/[sec(x)+tan(x)], then do a u-substitution. In the case of csc(x), multiply by [csc(x)+cot(x)]/[csc(x)+cot(x)], and do a u-substitution.
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