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    Find the integral of 1/(x^2+1)^2
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    (Original post by MEPS1996)
    Find the integral of 1/(x^2+1)^2
    What substitutions do you know that might help you? What would you do if the denominator were just (x^2 + 1)?
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    (Original post by davros)
    What substitutions do you know that might help you? What would you do if the denominator were just (x^2 + 1)?
    of course x=tan(t) substitution thanks
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    (Original post by MEPS1996)
    ...
    Alternatively, you could opt for the substitution x = \sinh u
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    (Original post by Khallil)
    Alternatively, you could opt for the substitution x = \sinh u
    Hey, I just wanted to ask a general question (couldn't find another FP2 thread).

    When finding the area of a cardioid, how do I find out what the limits are, directly from the r= equation?
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    (Original post by TheAsian)
    ...
    I'd presume so. It's been a while since I've looked at polar coordinates. Do you have an example with which I can work?

    Also, so that we don't derail the primary purpose of this thread, perhaps you could send me a VM (visitor message)?
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    (Original post by Khallil)
    I'd presume so. It's been a while since I've looked at polar coordinates. Do you have an example with which I can work?
    "find the area of a single loop of the curve with equation r = acos3x"

    The limits are pi/6 and -pi/6 but im not sure how they got that? thank you
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    (Original post by TheAsian)
    "find the area of a single loop of the curve with equation r = acos3x"

    The limits are pi/6 and -pi/6 but im not sure how they got that? thank you
    This is going to be a tough one to explain.

    Before looking at the polar curve, a very brief glance at the Cartesian trigonometric curve should give you a better idea of what is going on. For a loop to exist in the very first place, the endpoints of the region of the angle you're looking at must give a zero value for r. In this case, you could choose the region of x as the arrow I drew parallel to the x axis. Now it's just a matter of concerning yourself only with the positive values of r (as most exam boards don't bother with -ve values) for which the endpoints are equal to 0.



    (I couldn't get Wolfram to graph the output as [-a, a])
    In this respect, you could equally find the the integral \displaystyle \int_{\frac{\pi}{2}}^{\frac{5\pi  }{6}} and you'd get the same area.
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    (Original post by Khallil)
    This is going to be a tough one to explain.

    Before looking at the polar curve, a very brief glance at the trigonometric curve should give you a better idea of what is going on. For a loop to exist in the very first place, the endpoints of the region of the angle you're looking at must give a zero value for r. That much should be clear. Now it's just a matter of concerning yourself only with the positive values of r (as most exam boards don't bother with -ve values) for which the endpoints are equal to 0.



    (I couldn't get Wolfram to graph the output as [-a, a])
    In this respect, you could equally find the the integral \displaystyle \int_{\frac{\pi}{2}}^{\frac{5\pi  }{6}} and you'd get the same area.
    oh, okay! that's a lot clearer now, thank you vm!
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    (Original post by TheAsian)
    oh, okay! that's a lot clearer now, thank you vm!
    No prob. I edited a bit of my reply to make the "region of x for which the loop is enclosed" bit a tad clearer :-)
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    (Original post by Khallil)
    No prob. I edited a bit of my reply to make the "region of x for which the loop is enclosed" bit a tad clearer :-)
    it did clarify, thank you
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    (Original post by MEPS1996)
    Find the integral of 1/(x^2+1)^2
    I figured the question asked might be of some use to you too, so I posted my reply a few posts above.

    (Original post by TheAsian)
    it did clarify, thank you
    Nice question! Out of curiosity, did you get your final area as \frac{a^2 \pi}{12} ?
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    (Original post by Khallil)
    Nice question! Out of curiosity, did you get your final area as \frac{a^2 \pi}{12} ?
    yes! after finding the limits, it's really easy to do :rolleyes:
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    (Original post by TheAsian)
    yes! after finding the limits, it's really easy to do :rolleyes:
    Somebody's confident
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    (Original post by Khallil)
    Somebody's confident
    far from it, further maths is my scariest pursuit as of yet :afraid:
 
 
 
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