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\[\int\limits\limits_{-a}^{a} \sqrt{\frac{a-x}{a+x}}\]
Lets say, \( f(x)= \sqrt{\frac{a-x}{a+x}} \) , now f(X) is even or odd?
even or odd?

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Other answers:

i dont know :/@Fool
multiply and divide by the numerator
There is a property of definite integral which I think could be used here.
then the equation becomes solvable
hm letme see
hmm i think i know the answer, but not sure why it should be easy
let x=acosθ
can you post it ?
I was talking about this:
now you substitute x=acos O
answer is \[a\pi\] but not sure why yet
lol @sat
well i don't think this beast is even or odd
would be odd without the radical
ahh, i think dhashni has it!
No sat after you break it as dash showed, the first is even and the second is odd.
ahh, and second integral is 0 for sure
so the first term reduces to zero.
i got zero for the second one, maybe i am wrong
how second? the first is even and the second is odd.
no i am right, second on is zero
if even, then \[\int_{-a}^a f(x)dx=2\int_0^af(x)dx\]
Oh yes ... my apologies, yes the the second is zero.
For the first one we can use the step.
if odd then zero, and in any case we can actually compute the second integral right? \[\int\frac{x}{\sqrt{a^2-x^2}}dx=-\sqrt{a^2-x^2}\]
which is 0 and a and -a, so that one is gone.
So as the second one is odd and hence zero, we have to evaluate the first one.
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
sat, that's great :D
sorry, i meant the limits as x goes to a, not as x goes to zero
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
Lol OkaYy :D Thaanks a Lot :D

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