## Diyadiya 3 years ago Integration

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

2. FoolForMath

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. DHASHNI

multiply and divide by the numerator

6. FoolForMath

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

7. DHASHNI

then the equation becomes solvable

hm letme see

9. satellite73

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

10. Mr.crazzy

let x=acosθ

can you post it ?

12. DHASHNI

|dw:1326825738809:dw|

13. DHASHNI

|dw:1326825797080:dw|

14. FoolForMath

15. DHASHNI

now you substitute x=acos O

16. satellite73

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

17. rld613

lol @sat

18. satellite73

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

19. satellite73

would be odd without the radical

20. satellite73

ahh, i think dhashni has it!

21. FoolForMath

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

22. satellite73

ahh, and second integral is 0 for sure

23. FoolForMath

so the first term reduces to zero.

24. satellite73

hmm

25. satellite73

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

26. FoolForMath

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

27. satellite73

no i am right, second on is zero

28. satellite73

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

29. FoolForMath

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

30. FoolForMath

For the first one we can use the step.

31. satellite73

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. satellite73

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

33. FoolForMath

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

34. satellite73

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. FoolForMath

sat, that's great :D

36. satellite73

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

37. satellite73

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