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Math Help - Trig integral.

  1. #1
    Member Marconis's Avatar
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    Trig integral.

    We are learning summation currently, but since he is not finished with the lesson he assigned us a few more substitution problems. I am finding them very difficult, actually.

    \int \sec{2\theta} \tan{2\theta} d\theta

    I let u= \sec{2\theta} to get du= (\sec{2\theta} \tan{2\theta}) 2 d\theta

    All that's missing in the original problem is "2". I have never encountered a problem where taking the derivative of u would substitute the entire problem, with the exception of 2, of course. The only thing I'd get out of it is \frac{1}{2} + C, and that doesn't seem right.

    Next problem is also causing me some difficulty...

    \int \frac{cos(\pi/x)}{x^2} dx

    I let u= \frac{\pi}{x} to get du= \frac{-\pi}{x^2} dx

    Final answer \frac{-1}{\pi} \sin{(pi/x)} + C

    I have a huge hunch I did that one wrong.

    Any assistance is greatly appreciated, as always.
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  2. #2
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    \displaystyle \int \sec(2t)\tan(2t) \, dt<br />

    let u = 2t

    du = 2 \, dt

    \displaystyle \frac{1}{2} \int \sec(2t)\tan(2t)  \cdot 2 \, dt

    \displaystyle \frac{1}{2} \int \sec(u)\tan(u)  \, du = \frac{1}{2} \sec(u) + C

    back sub and finish.


    check the 2nd problem by taking the derivative of your solution.
    Last edited by mr fantastic; February 13th 2011 at 04:47 PM.
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  3. #3
    Member Marconis's Avatar
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    Thanks for your help.

    Sorry about the bump. I'm on so many forums, and at times it is hard to remember the rules of each. I read them over and will take great care to follow them from now on.


    Thanks also for the thread move, Mod. My apologies for that as well.

    Anyway, back on track. I hate that I fell victim to over-thinking that first problem. The reason I didn't think to do that is because of the derivative of \sec{x} equaling  \tan{x} + \sec{x}. The lack of the plus sign threw me off. It's clear now, though.

    Well now for that second one, I'm actually having difficulty taking the derivative of my solution...Could the reason be that I did the integral incorrectly?
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  4. #4
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    \dfrac{d}{dx} \left[-\dfrac{1}{\pi} \sin\left(\dfrac{\pi}{x}\right)\right]

    chain rule ...

    -\dfrac{1}{\pi} \cos\left(\dfrac{\pi}{x}\right) \cdot \left[-\dfrac{\pi}{x^2}\right]

    \dfrac{\cos\left(\dfrac{\pi}{x}\right)}{x^2}


    you really need to practice ... at this stage of studying the calculus, basic derivatives should be second nature.
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  5. #5
    Member Marconis's Avatar
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    Well, good to know I was right. As for practicing derivatives, I think I am pretty good with them because I have been able to do them for all these integration problems without any issue. Seeing pi throws me off sometimes, and it's been a long day of studying so I drew a blank.

    Thanks for the help.
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