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FP2 Integral

Find the integral of 1/(x^2+1)^2
Reply 1
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)?
Reply 2
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
Original post by MEPS1996
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Alternatively, you could opt for the substitution x=sinhux = \sinh u
Reply 4
Original post by Khallil
Alternatively, you could opt for the substitution x=sinhux = \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?
Original post by TheAsian
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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)?
(edited 10 years ago)
Reply 6
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
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 rr. In this case, you could choose the region of xx as the arrow I drew parallel to the xx axis. Now it's just a matter of concerning yourself only with the positive values of rr (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][-a, a])
In this respect, you could equally find the the integral π25π6\displaystyle \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} and you'd get the same area.
(edited 10 years ago)
Reply 8
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 rr. That much should be clear. Now it's just a matter of concerning yourself only with the positive values of rr (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][-a, a])
In this respect, you could equally find the the integral π25π6\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!
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 :-)
Reply 10
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:smile:
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:smile:


Nice question! Out of curiosity, did you get your final area as a2π12\frac{a^2 \pi}{12} ?
(edited 10 years ago)
Reply 12
Original post by Khallil
Nice question! Out of curiosity, did you get your final area as a2π12\frac{a^2 \pi}{12} ?


yes! after finding the limits, it's really easy to do :rolleyes:
Original post by TheAsian
yes! after finding the limits, it's really easy to do :rolleyes:


Somebody's confident :wink:
Reply 14
Original post by Khallil
Somebody's confident :wink:

far from it, further maths is my scariest pursuit as of yet :afraid:

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