Integration

- Diyadiya

Integration

- schrodinger

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- Diyadiya

\[\int\limits\limits_{-a}^{a} \sqrt{\frac{a-x}{a+x}}\]

- anonymous

Lets say, \( f(x)= \sqrt{\frac{a-x}{a+x}} \) , now f(X) is even or odd?

- Diyadiya

even or odd?

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## More answers

- Diyadiya

i dont know :/@Fool

- anonymous

multiply and divide by the numerator

- anonymous

There is a property of definite integral which I think could be used here.

- anonymous

then the equation becomes solvable

- Diyadiya

hm letme see

- anonymous

hmm i think i know the answer, but not sure why it should be easy

- anonymous

let x=acosθ

- Diyadiya

can you post it ?

- anonymous

|dw:1326825738809:dw|

- anonymous

|dw:1326825797080:dw|

- anonymous

I was talking about this: http://en.wikibooks.org/wiki/Calculus/Definite_integral#Even_and_odd_functions

- anonymous

now you substitute x=acos O

- anonymous

answer is
\[a\pi\] but not sure why yet

- anonymous

lol @sat

- anonymous

well i don't think this beast is even or odd

- anonymous

would be odd without the radical

- anonymous

ahh, i think dhashni has it!

- anonymous

No sat after you break it as dash showed, the first is even and the second is odd.

- anonymous

ahh, and second integral is 0 for sure

- anonymous

so the first term reduces to zero.

- anonymous

hmm

- anonymous

i got zero for the second one, maybe i am wrong

- anonymous

how second? the first is even and the second is odd.

- anonymous

no i am right, second on is zero

- anonymous

if even, then
\[\int_{-a}^a f(x)dx=2\int_0^af(x)dx\]

- anonymous

Oh yes ... my apologies, yes the the second is zero.

- anonymous

For the first one we can use the step.

- anonymous

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

which is 0 and a and -a, so that one is gone.

- anonymous

So as the second one is odd and hence zero, we have to evaluate the first one.

- anonymous

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

- anonymous

sat, that's great :D

- anonymous

sorry, i meant the limits as x goes to a, not as x goes to zero

- anonymous

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

Lol OkaYy :D Thaanks a Lot :D

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