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\[\int\limits _0^1\frac{1}{\alpha-\cos x}dx=\frac{\pi}{\sqrt{\alpha^2-1}}\]
i tried \[\tan \frac{x}{2}=u\] i have \[\huge \int\limits \frac{1}{\alpha-\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=2\huge \int\limits \frac{1}{\alpha-1+(1+\alpha)t^2}dt\]
\[\huge \frac{2}{\alpha+1}\int\limits \frac{dt}{(\sqrt{\frac{\alpha-1}{\alpha+1}})^2+t^2}=\frac{2}{\alpha+1}\tan^{-1}(\frac{t}{\sqrt{\frac{\alpha-1}{\alpha+1}}})\]

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Other answers:

first of all tan pi/2 is undefined,is there anything wrong here?
oh something missing the whole thing shud be divided by \[\sqrt{\frac{\alpha-1}{\alpha+1}}\]
\[\huge \frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\frac{\frac{x}{2}}{\sqrt{\frac{\alpha-1}{\alpha+1}}}\]
Did you mutliply by conjugate?
Wait, you're proving. My bad, I thought you were solving using trig sub.
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
Did you get anything like that?
|dw:1363452681715:dw| i thought thig is +
Yeah, it should.
Did you get anything like that?
yes thats my first solution there on top instead of z i used t
  • 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
  • phi
Here are the gory details
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these makes it clear ,thanks but how do we simplify to get the final answer
initially \[\int _0^\pi\]
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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
  • 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
  • phi
Here is the problem using the correct limits
1 Attachment
thanks again for helping sry for the error

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