## Diyadiya 4 years ago Integration

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

2. anonymous

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

even or odd?

i dont know :/@Fool

5. anonymous

multiply and divide by the numerator

6. anonymous

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

7. anonymous

then the equation becomes solvable

hm letme see

9. anonymous

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

10. anonymous

let x=acosθ

can you post it ?

12. anonymous

|dw:1326825738809:dw|

13. anonymous

|dw:1326825797080:dw|

14. anonymous

15. anonymous

now you substitute x=acos O

16. anonymous

answer is $a\pi$ but not sure why yet

17. anonymous

lol @sat

18. anonymous

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

19. anonymous

would be odd without the radical

20. anonymous

ahh, i think dhashni has it!

21. anonymous

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

22. anonymous

ahh, and second integral is 0 for sure

23. anonymous

so the first term reduces to zero.

24. anonymous

hmm

25. anonymous

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

26. anonymous

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

27. anonymous

no i am right, second on is zero

28. anonymous

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

29. anonymous

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

30. anonymous

For the first one we can use the step.

31. anonymous

if odd then zero, and in any case we can actually compute the second integral right? $\int\frac{x}{\sqrt{a^2-x^2}}dx=-\sqrt{a^2-x^2}$

32. anonymous

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

33. anonymous

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

34. 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

35. anonymous

sat, that's great :D

36. anonymous

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

37. 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