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## Jonask Group Title prove intergral one year ago one year ago

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1. Jonask

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

2. Jonask

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. Jonask

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

4. Jonask

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

5. Jonask

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

6. Jonask

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

7. abb0t

Did you mutliply by conjugate?

8. Jonask

yes

9. linshan789

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

10. Jonask

|dw:1363451580417:dw|

11. Jonask

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

12. Jonask

my problem here is tan pi/2 is undefined

13. Jonask

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. Jonask

last denominator cos^2x

15. linshan789

|dw:1363445220695:dw|

16. linshan789

Did you get anything like that?

17. Jonask

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

18. linshan789

Yeah, it should.

19. linshan789

Did you get anything like that?

20. Jonask

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. Jonask

can you explain how you did it

23. phi

Here are the gory details

24. Jonask

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

25. Jonask

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

26. Jonask

initially $\int _0^\pi$

27. Jonask

28. Jonask

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

29. Jonask

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. Jonask

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

31. Jonask

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

aha eureka!! thanks guys

34. phi

Here is the problem using the correct limits

35. Jonask

thanks again for helping sry for the error