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Math Help - Integration problem

  1. #1
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    Integration problem

    I'm having trouble integrating this function:

    \int \sqrt{2-2\cos{4t}} dt

    So far I've tried substitution and that didn't work out so well. Please help!
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  2. #2
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    Quote Originally Posted by RU2010 View Post
    I'm having trouble integrating this function:

    \int \sqrt{2-2\cos{4t}} dt

    So far I've tried substitution and that didn't work out so well. Please help!
    Note that \cos(4t) = 2 \cos^2 (2t) - 1. See what you can do with that.
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  3. #3
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    Ok, I made the double angle substitution, but I'm still drawing a blank on how to get this one started. Which method of integration should I use?
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  4. #4
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    Better yet, note that \cos{(4t)} = 1 - 2\sin^2{(2t)}.
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  5. #5
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    Ok, I've used the trig identities to change the integral to this:

    \int \sqrt{4 \sin^2{2t}}dt

    I think what is throwing me off is the fact that this function is under the radial. At this point, would substitution work if I use u= \sin^2{2t}? Is this the correct integration method to use?
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  6. #6
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    Scrap that substition, it will not work. Which method of integration do I need to look at?
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  7. #7
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    I think I might have found the solution. The problem is actually a definite integral problem with the lower limit of integration as 0 and the upper limit of integration as pi/2.

    I've worked the problem out and came to a solution of -2. Is this correct?
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  8. #8
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    \int \sqrt{4 \sin^2{2t}}dt

    \int \sqrt{2^2 \sin^2{2t}}dt

    \int \sqrt{(2 \sin{2t})^2}dt

    \int 2 \sin{2t}dt

    Its a gimmie from here!
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