## Jonask one year ago prove intergral

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

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

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

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

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

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

7. abb0t

Did you mutliply by conjugate?

yes

9. linshan789

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

|dw:1363451580417:dw|

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

my problem here is tan pi/2 is undefined

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

last denominator cos^2x

15. linshan789

|dw:1363445220695:dw|

16. linshan789

Did you get anything like that?

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

18. linshan789

Yeah, it should.

19. linshan789

Did you get anything like that?

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

can you explain how you did it

23. phi

Here are the gory details

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

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

initially $\int _0^\pi$

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

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

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

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

aha eureka!! thanks guys

34. phi

Here is the problem using the correct limits