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

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

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

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

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

Jonask
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0Did you mutliply by conjugate?

linshan789
 one year ago
Best ResponseYou've already chosen the best response.0Wait, you're proving. My bad, I thought you were solving using trig sub.

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0same thing if you can solve it then you actually proved it

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0my problem here is tan pi/2 is undefined

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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0last denominator cos^2x

linshan789
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363445220695:dw

linshan789
 one year ago
Best ResponseYou've already chosen the best response.0Did you get anything like that?

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363452681715:dw i thought thig is +

linshan789
 one year ago
Best ResponseYou've already chosen the best response.0Did you get anything like that?

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

phi
 one year 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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0can you explain how you did it

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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\pi}{\sqrt{\alpha^21}}\]

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0initially \[\int _0^\pi\]

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0i had typed the wrong boundaries 0 to 1 instead of 0 to pi

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

Jonask
 one year 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

Jonask
 one year 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
 one year 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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0aha eureka!! thanks guys

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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0thanks again for helping sry for the error
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