prove intergral

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prove intergral

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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?
yes
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
|dw:1363445220695:dw|
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
\[\frac{\pi}{\sqrt{\alpha^2-1}}\]
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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