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Diyadiya
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\[\int\limits\limits_{-a}^{a} \sqrt{\frac{a-x}{a+x}}\]
FoolForMath
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Lets say, \( f(x)= \sqrt{\frac{a-x}{a+x}} \) , now f(X) is even or odd?
Diyadiya
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even or odd?
Diyadiya
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i dont know :/@Fool
DHASHNI
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multiply and divide by the numerator
FoolForMath
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There is a property of definite integral which I think could be used here.
DHASHNI
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then the equation becomes solvable
Diyadiya
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hm letme see
anonymous
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hmm i think i know the answer, but not sure why it should be easy
Mr.crazzy
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let x=acosθ
Diyadiya
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can you post it ?
DHASHNI
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|dw:1326825738809:dw|
DHASHNI
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|dw:1326825797080:dw|
DHASHNI
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now you substitute x=acos O
anonymous
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answer is
\[a\pi\] but not sure why yet
rld613
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lol @sat
anonymous
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well i don't think this beast is even or odd
anonymous
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would be odd without the radical
anonymous
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ahh, i think dhashni has it!
FoolForMath
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No sat after you break it as dash showed, the first is even and the second is odd.
anonymous
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ahh, and second integral is 0 for sure
FoolForMath
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so the first term reduces to zero.
anonymous
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hmm
anonymous
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i got zero for the second one, maybe i am wrong
FoolForMath
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how second? the first is even and the second is odd.
anonymous
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no i am right, second on is zero
anonymous
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if even, then
\[\int_{-a}^a f(x)dx=2\int_0^af(x)dx\]
FoolForMath
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Oh yes ... my apologies, yes the the second is zero.
FoolForMath
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For the first one we can use the step.
anonymous
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if odd then zero, and in any case we can actually compute the second integral right?
\[\int\frac{x}{\sqrt{a^2-x^2}}dx=-\sqrt{a^2-x^2}\]
anonymous
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which is 0 and a and -a, so that one is gone.
FoolForMath
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So as the second one is odd and hence zero, we have to evaluate the first one.
anonymous
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and it is somewhat of a pain because it is an improper integral, but you will get
\[a\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})\] and then taking the limit as x goes to 0 you get the limit as x goes to zero of arctan(x) =pi/2 from one direction, -pi/2 from other direction subtract and get pi and that explains the answer of
\[a\pi\] whrew
FoolForMath
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sat, that's great :D
anonymous
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sorry, i meant the limits as x goes to a, not as x goes to zero
anonymous
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x goes to a,
\[\frac{x}{\sqrt{a^2-x^2}}\] goes to infinity and
\[\arctan(\infty)\] goes to
\[\frac{\pi}{2}\] is what i meant
Diyadiya
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Lol OkaYy :D Thaanks a Lot :D