## anonymous 3 years ago prove intergral

1. anonymous

$\int\limits _0^1\frac{1}{\alpha-\cos x}dx=\frac{\pi}{\sqrt{\alpha^2-1}}$

2. anonymous

i tried $\tan \frac{x}{2}=u$ i have $\huge \int\limits \frac{1}{\alpha-\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=2\huge \int\limits \frac{1}{\alpha-1+(1+\alpha)t^2}dt$

3. anonymous

$\huge \frac{2}{\alpha+1}\int\limits \frac{dt}{(\sqrt{\frac{\alpha-1}{\alpha+1}})^2+t^2}=\frac{2}{\alpha+1}\tan^{-1}(\frac{t}{\sqrt{\frac{\alpha-1}{\alpha+1}}})$

4. anonymous

first of all tan pi/2 is undefined,is there anything wrong here?

5. anonymous

oh something missing the whole thing shud be divided by $\sqrt{\frac{\alpha-1}{\alpha+1}}$

6. anonymous

$\huge \frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\frac{\frac{x}{2}}{\sqrt{\frac{\alpha-1}{\alpha+1}}}$

7. abb0t

Did you mutliply by conjugate?

8. anonymous

yes

9. anonymous

Wait, you're proving. My bad, I thought you were solving using trig sub.

10. anonymous

|dw:1363451580417:dw|

11. anonymous

same thing if you can solve it then you actually proved it

12. anonymous

my problem here is tan pi/2 is undefined

13. anonymous

METHOD II RATIONALISE AND SEPERATE $\int\limits \frac{1}{\alpha-\cos x}dx=\int\limits \frac{\alpha+\cos x}{\alpha^2-\cos^2x}dx=\int\limits \frac{\alpha}{\alpha^2-\cos^2x}+\int\limits \frac{\cos x}{\alpha^2-\cos x}$

14. anonymous

last denominator cos^2x

15. anonymous

|dw:1363445220695:dw|

16. anonymous

Did you get anything like that?

17. anonymous

|dw:1363452681715:dw| i thought thig is +

18. anonymous

Yeah, it should.

19. anonymous

Did you get anything like that?

20. anonymous

yes thats my first solution there on top instead of z i used t

21. phi

I got $\int\limits _0^1\frac{1}{\alpha-\cos x}dx= \frac{2}{\sqrt{\alpha^2-1}} \tan^{-1}\left(\sqrt{\frac{\alpha+1}{\alpha-1}}\tan\frac{1}{2}\right)$ I don't see the inverse tan giving pi/2 except at alpha->1

22. anonymous

can you explain how you did it

23. phi

Here are the gory details

24. anonymous

these makes it clear ,thanks but how do we simplify to get the final answer

25. anonymous

$\frac{\pi}{\sqrt{\alpha^2-1}}$

26. anonymous

initially $\int _0^\pi$

27. anonymous

28. anonymous

i had typed the wrong boundaries 0 to 1 instead of 0 to pi

29. anonymous

SO THE PROBLEM IS HERE $\frac{1}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (x/2)}\right)$

30. anonymous

$\frac{1}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (\pi/2)}\right)$ the part inside is undefined

31. anonymous

$\frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (\pi/2)}\right)$ i mean 2

32. phi

that makes more sense, because I was going to say the original expression is false. however, for this new limit, you can say tan(x) as x-> pi/2 approaches infinity and the inverse tangent of x--> infinity approaches pi/2

33. anonymous

aha eureka!! thanks guys

34. phi

Here is the problem using the correct limits

35. anonymous

thanks again for helping sry for the error

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