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Math Help - Trig Sub to Integration by Parts (Supposedly)

  1. #1
    Member VitaX's Avatar
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    Trig Sub to Integration by Parts (Supposedly)

    L = \int_{0}^{\frac{\pi}{4}} 2\sqrt{1+t^2} dt \rightarrow 2\int_{0}^{\frac{\pi}{4}} \sqrt{1^2 +t^2} dt

    t = tan(\theta) \rightarrow dt = sec^2(\theta) d\theta

    = 2 \int \sqrt{1 + tan^2(\theta)} * sec^2(\theta) d\theta \rightarrow 2\int sec^3(\theta) d\theta

    = 2 \int (1+tan^2(\theta)) * sec(\theta) d\theta \rightarrow 2\int sec(\theta) + sec(\theta)tan^2(\theta) d\theta

    =2\int sec(\theta) d\theta + 2\int sec(\theta)tan^2(\theta) d\theta

    =2 \int sec(\theta) d\theta \rightarrow 2\ln[sec(\theta) + tan(\theta)]

    To simplify the first integral answer isn't it just setting up a right triangle and figuring out what the values will be? I got sec(\theta) = \sqrt{t^2 +1} and tan(\theta) = t

    So then the answer to that first integral would be 2\ln[\sqrt{t^2+1} + t] right? To be honest I'm really lost on this entire problem after I got half way through the integral. How do I finish this problem up exactly? And when exactly do I change the limits on this? Assuming that I will have to at some point. I attached what wolfram got as an answer as well for the indefinite integral for a reference. I'm not reallly familiar with the reduction formula as I never learned that in calculus 2. Perhaps I also went too far in the integral above after looking at wolfram's work, though I'm not too sure again. Any help appreciated.

    *Only put "Supposedly" in title because I wasn't completely sure if integration by parts is in this problem or not. But after submitting it seems more likely that it is.
    Attached Thumbnails Attached Thumbnails Trig Sub to Integration by Parts (Supposedly)-integral.jpg  
    Last edited by VitaX; October 8th 2010 at 01:31 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    \displaystyle \int \sec^3 \theta d\theta

    \displaystyle \int \sec^3 \theta d\theta = \int \sec \theta d\theta + \int \sec \theta \tan \theta (\tan \theta ) d \theta

    the second integral by parts

    let dv be sec tan and u tan so we will have

    \displaystyle \int \sec \theta \tan \theta ( \tan \theta ) d\theta = \sec \theta \tan \theta - \int \sec \theta (\sec ^2 \theta ) d\theta

    \displaystyle \int \sec ^3 \theta d\theta = \int \sec \theta d\theta + \sec \theta \tan \theta - \int \sec ^3 \theta

    \displaystyle 2 \int sec ^3 \theta d\theta = \int \sec \theta d\theta+ \sec \theta \tan \theta

    how to integrate \displaystyle sec \theta here multiply the denominator and the nominator by \sec \theta + \tan \theta

    \displaystyle  \int \sec \theta d\theta = \int \frac{\sec \theta (\sec \theta + \tan \theta)}{\sec \theta + \tan \theta} d\theta

    \displaystyle \int \frac{\sec ^2 \theta + \sec \theta \tan \theta }{\sec \theta + \tan \theta } d\theta

    the nominator is the derivative of denominator so ln(denominator)
    since \displaystyle \left(ln (f) \right)' = \frac{f'}{f}

    about the limits if u write ur integral result using t variable there is no need to change the limits since t limits u have it but if u do not write the result using t variable ( theta variable ) then u need to find the limits of theta
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  3. #3
    Member VitaX's Avatar
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    I understand your work for the integration by parts. But what I'm having trouble with is accumulating an answer from the work in your post and what I came up with for the integral of the first part in my post. What would be the final answer to the integral before finding the length of the curve using the limits provided? Is my answer at least correct in my original post for the first integral?
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  4. #4
    MHF Contributor Amer's Avatar
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    yeah it is correct just sub the limits
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  5. #5
    Member VitaX's Avatar
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    Sorry, but what I'm asking is I don't understand how to write out the full answer to the integral. I understand that after I found the integral of the first part of this:

    2\int sec(\theta) d\theta + 2 \int sec(\theta)tan(\theta)tan(\theta) d\theta

    2\int sec(\theta) = 2\ln[sec(\theta) + tan(\theta)]

    And from the right triangle I get sec(\theta) = \sqrt{t^2 +1} ; tan(\theta) = t

    So from there I say the following is my answer to the first integral \Longrightarrow 2\ln[\sqrt{t^2 + 1} + t]

    And I see this part in the wolfram answer but they do not have a 2 in front as a coefficient so I'm not too sure what is up with that. But what I'm really wondering about is how to finish the rest of this answer so I can start plugging in the limits for t and get an answer for the length of the curve. I was trying to make sense of your work but, I can't depict an answer from that to add to the one above. That's my question, if you or someone else can help with that it would be great. Or maybe walk me through the steps again as it's a bit confusing to me. Either way that is what I'm stuck on. I hope I clarified my 2nd post on this thread a bit more anyway.
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  6. #6
    MHF Contributor Amer's Avatar
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    ok I will solve step by step if that what u want we want to integrate

    \displaystyle \int _{0}^{\frac{\pi}{4}} 2 \sqrt{t^2 +1} dt

    lets first solve the integral without the limits

    \displaystyle \int 2 \sqrt{t^2 +1 } dt

    sub \displaystyle \tan \theta = t we will have the follow integral

    \displaystyle \int 2 \sec ^3 \theta \; d\theta = \int 2 \sec \theta d\theta  + \int 2 \sec \theta \tan ^2 \theta d\theta

    \displaystyle \int 2 \sec ^3 \theta \; d\theta = 2 \int \sec \theta \; d\theta  + 2 \int \sec \theta \tan ^2 \theta \; d\theta

    second integral by parts as I said before

    \displaystyle \int 2 \sec ^3 \theta \; d\theta = 2 \int \sec \theta d\theta + 2\left( \tan \theta \sec \theta - \int \sec ^3 \theta d\theta \right)

    \displaystyle \int 2 \sec ^3 \; d\theta  = 2 \int \sec \theta \; d\theta + 2 \sec \theta \tan \theta - 2 \int \sec ^3 \theta \; d\theta

    as u can see in the right hand side we have our integral so

    \displaystyle 2\int \sec ^3 d\theta + 2\int \sec ^3 \theta d\theta = 2 \int \sec \theta d\theta + 2 \sec \theta \tan \theta

    so

    \displaystyle 4 \int \sec ^3 d\theta = 2\ln \mid \sec \theta + \tan \theta \mid +2 \sec \theta \tan \theta

    but we want to integrate \displaystyle \int 2 \sec\theta \; d\theta
    divide both sides by 2

    \displaystyle 2 \int \sec ^3 \theta \; d\theta = \ln \mid \sec \theta + \tan \theta \mid + \sec \theta \tan \theta +C

    so

    \displaystyle \int 2 \sqrt{t^2 +1 } \; dt = \ln \mid \sqrt{t^2 +1} + t \mid + \sqrt{t^2 +1} \; t +C

    in the End

    \displaystyle \int _{0}^{\frac{\pi}{4}} 2 \sqrt{t^2 +1} dt = \left[ \ln (\sqrt{t^2+1} + t ) + t \sqrt{t^2+1} \right]_{0}^{\frac{\pi}{4}}

    hope u understand it, something unclear ??
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  7. #7
    Member VitaX's Avatar
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    Thanks a lot. I happened around the integral of Sec^3(\theta) on wikipedia luckily, so I did the integral a little bit ago and I did come up with your answer. Also did it a little bit differently than you I guess but overall got the same answer at the end. I guess what I wasn't seeing in your earlier post was that its integration by parts matching with the original integral. But now that I see that it is, things are much clearer now. My overall answer for the length was 3.4538 units.
    Last edited by VitaX; October 8th 2010 at 09:23 PM.
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