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anonymous
 3 years ago
prove intergral
anonymous
 3 years ago
prove intergral

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits _0^1\frac{1}{\alpha\cos x}dx=\frac{\pi}{\sqrt{\alpha^21}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i tried \[\tan \frac{x}{2}=u\] i have \[\huge \int\limits \frac{1}{\alpha\frac{1t^2}{1+t^2}}\frac{2dt}{1+t^2}=2\huge \int\limits \frac{1}{\alpha1+(1+\alpha)t^2}dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{2}{\alpha+1}\int\limits \frac{dt}{(\sqrt{\frac{\alpha1}{\alpha+1}})^2+t^2}=\frac{2}{\alpha+1}\tan^{1}(\frac{t}{\sqrt{\frac{\alpha1}{\alpha+1}}})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first of all tan pi/2 is undefined,is there anything wrong here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh something missing the whole thing shud be divided by \[\sqrt{\frac{\alpha1}{\alpha+1}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{2}{\sqrt{\alpha^21}}\tan^{1}\frac{\frac{x}{2}}{\sqrt{\frac{\alpha1}{\alpha+1}}}\]

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0Did you mutliply by conjugate?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait, you're proving. My bad, I thought you were solving using trig sub.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363451580417:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0same thing if you can solve it then you actually proved it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my problem here is tan pi/2 is undefined

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0METHOD II RATIONALISE AND SEPERATE \[\int\limits \frac{1}{\alpha\cos x}dx=\int\limits \frac{\alpha+\cos x}{\alpha^2\cos^2x}dx=\int\limits \frac{\alpha}{\alpha^2\cos^2x}+\int\limits \frac{\cos x}{\alpha^2\cos x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0last denominator cos^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363445220695:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you get anything like that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363452681715:dw i thought thig is +

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you get anything like that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes thats my first solution there on top instead of z i used t

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I got \[ \int\limits _0^1\frac{1}{\alpha\cos x}dx= \frac{2}{\sqrt{\alpha^21}} \tan^{1}\left(\sqrt{\frac{\alpha+1}{\alpha1}}\tan\frac{1}{2}\right)\] I don't see the inverse tan giving pi/2 except at alpha>1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you explain how you did it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0these makes it clear ,thanks but how do we simplify to get the final answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\pi}{\sqrt{\alpha^21}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0initially \[\int _0^\pi\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i had typed the wrong boundaries 0 to 1 instead of 0 to pi

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0SO THE PROBLEM IS HERE \[\frac{1}{\sqrt{\alpha^21}}\tan^{1}\left( {\sqrt{\frac{\alpha1}{\alpha+1}}\tan (x/2)}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{\sqrt{\alpha^21}}\tan^{1}\left( {\sqrt{\frac{\alpha1}{\alpha+1}}\tan (\pi/2)}\right)\] the part inside is undefined

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{2}{\sqrt{\alpha^21}}\tan^{1}\left( {\sqrt{\frac{\alpha1}{\alpha+1}}\tan (\pi/2)}\right)\] i mean 2

phi
 3 years ago
Best ResponseYou've already chosen the best response.0that 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0aha eureka!! thanks guys

phi
 3 years ago
Best ResponseYou've already chosen the best response.0Here is the problem using the correct limits

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks again for helping sry for the error
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