Jonask Group Title prove intergral one year ago one year ago

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

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 Group Title

Did you mutliply by conjugate?

yes

9. linshan789 Group Title

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 Group Title

|dw:1363445220695:dw|

16. linshan789 Group Title

Did you get anything like that?

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

18. linshan789 Group Title

Yeah, it should.

19. linshan789 Group Title

Did you get anything like that?

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

21. phi Group Title

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 Group Title

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 Group Title

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 Group Title

Here is the problem using the correct limits