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Math Help - Definite integral.

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

    This one was a tad tricky for me, let me know if I am incorrect. I apologize in advance in that I do not know how to add numbers to the integral sign, couldn't find it on Latex.

    Interval is [1/6] to [1/2]

    \int \csc{\pi}t \cot{\pi}t dt

    Let u=\pi t

    du=\pi dt

    \frac{1}{\pi} \int \csc{(u)} \cot{(u)} du

    \frac{1}{\pi} \int -\csc{(u)}

    [ \frac{1}{\pi} \ast (-\csc{(\frac{1}{2}})) ] - [ \frac{1}{\pi} \ast (-\csc{(\frac{1}{6}})) ]

    Answer: -.1133

    Thanks for looking it over.
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  2. #2
    A Plied Mathematician
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    You have two options when doing a u substitution on a definite integral: you can either change the limits to the new variable, or, when you've computed the anti-derivative, change it back to the old variable and plug in the old limits. You made the mistake of plugging in the old limits with the u variable, which is like comparing apples to oranges. Do you follow me?
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  3. #3
    Member Marconis's Avatar
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    Oh man I just realized what I've done! I forgot that you're supposed to plug those values in where it is a variable, not where it is u! That was a bad slip. I'll redo it after I eat lunch.

    By a mistake I treated u as the variable, when I was supposed to put pi t back in and THEN plug in the two numbers.
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  4. #4
    A Plied Mathematician
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    Just do this:

    u(1/6)=\pi/6, and

    u(1/2)=\pi/2. Plug those in thus:

    \dfrac{1}{\pi}\left[-\csc(\pi/2)-(-\csc(\pi/6))\right].
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  5. #5
    Member Marconis's Avatar
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    *Edit- Sorry I was typing that all up as you replied*

    \int \csc{\pi}t \cot{\pi}t dt

    Let u=\pi t

    du=\pi dt

    \frac{1}{\pi} \int \csc{(u)} \cot{(u)} du

    \frac{1}{\pi} \int -\csc{(u)}

    [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{2}))} ] - [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{6}))} ]

    Is this right, now?
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  6. #6
    A Plied Mathematician
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    Correct. Of course, you can simplify a bit...
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  7. #7
    Member Marconis's Avatar
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    Yeah I do things the longer way at first. Thanks so much for your help, Ackbeet. Glad I did it correctly.
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  8. #8
    A Plied Mathematician
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    You're welcome. Have a good one!
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  9. #9
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    Quote Originally Posted by Marconis View Post
    *Edit- Sorry I was typing that all up as you replied*

    \int \csc{\pi}t \cot{\pi}t dt

    Let u=\pi t

    du=\pi dt

    \frac{1}{\pi} \int \csc{(u)} \cot{(u)} du

    \frac{1}{\pi} \int -\csc{(u)}
    But you should not have the integral sign there. You have already done the integral. What you should have is either \frac{1}{\pi} (-csc(u))+ C
    and then evaluate- the "C" will cancel

    or, better, \left[\frac{1}{\pi} (-csc(u)) \right]_{u= \pi/6}^{\pi/2}

    [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{2}))} ] - [\frac{1}{\pi} \ast (-\csc{(\pi\frac{1}{6}))} ]

    Is this right, now?
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  10. #10
    Member Marconis's Avatar
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    I need to break that habit. I always do that.

    So guys...I'm trying to find the answer and my calculator keeps saying ERROR: DOMAIN. I've tried typing in Ackbeets way and my way, and it keeps happening.
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  11. #11
    A Plied Mathematician
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    Hmm. Let's clean up the notation a bit and see if that doesn't help out in terms of evaluating on the calculator:

    \displaystyle-\frac{1}{\pi} \csc\left(\frac{\pi}{2}\right) + \frac{1}{\pi} \csc\left(\frac{\pi}{6}\right)=\frac{1}{\pi}\left(  \csc\left(\frac{\pi}{6}\right)-\csc\left(\frac{\pi}{2}\right)\right)=

    \displaystyle\frac{1}{\pi}\left(\frac{1}{\sin(\pi/6)}-\frac{1}{\sin(\pi/2)}\right).

    Make sure your calculator is in radian mode, and try evaluating that. What kind of calculator do you have?
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  12. #12
    Member Marconis's Avatar
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    I'm getting .31831. Is that what you got?

    I am using a TI-83. I prefer the TI-89 but our dept. doesn't let us use it.
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  13. #13
    A Plied Mathematician
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    Yep, that's what I get. And it's correct. See here.

    So the "answer" posted in the OP is wrong.
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  14. #14
    Member Marconis's Avatar
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    THAT SITE IS AWESOME! Thanks! Now I can check all my answers on there.
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  15. #15
    A Plied Mathematician
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    Yep. WolframAlpha is a beaut. You're welcome. Have a good one!
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