## Diyadiya Group Title Integration 2 years ago 2 years ago

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

2. FoolForMath Group Title

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 Group Title

multiply and divide by the numerator

6. FoolForMath Group Title

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

7. DHASHNI Group Title

then the equation becomes solvable

hm letme see

9. satellite73 Group Title

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

10. Mr.crazzy Group Title

let x=acosθ

can you post it ?

12. DHASHNI Group Title

|dw:1326825738809:dw|

13. DHASHNI Group Title

|dw:1326825797080:dw|

14. FoolForMath Group Title

15. DHASHNI Group Title

now you substitute x=acos O

16. satellite73 Group Title

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

17. rld613 Group Title

lol @sat

18. satellite73 Group Title

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

19. satellite73 Group Title

would be odd without the radical

20. satellite73 Group Title

ahh, i think dhashni has it!

21. FoolForMath Group Title

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

22. satellite73 Group Title

ahh, and second integral is 0 for sure

23. FoolForMath Group Title

so the first term reduces to zero.

24. satellite73 Group Title

hmm

25. satellite73 Group Title

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

26. FoolForMath Group Title

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

27. satellite73 Group Title

no i am right, second on is zero

28. satellite73 Group Title

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

29. FoolForMath Group Title

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

30. FoolForMath Group Title

For the first one we can use the step.

31. satellite73 Group Title

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 Group Title

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

33. FoolForMath Group Title

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

34. satellite73 Group Title

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 Group Title

sat, that's great :D

36. satellite73 Group Title

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

37. satellite73 Group Title

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