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

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

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

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

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

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

abb0tBest ResponseYou've already chosen the best response.0
Did you mutliply by conjugate?
 one year ago

linshan789Best ResponseYou've already chosen the best response.0
Wait, you're proving. My bad, I thought you were solving using trig sub.
 one year ago

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

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

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

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

linshan789Best ResponseYou've already chosen the best response.0
dw:1363445220695:dw
 one year ago

linshan789Best ResponseYou've already chosen the best response.0
Did you get anything like that?
 one year ago

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

linshan789Best ResponseYou've already chosen the best response.0
Did you get anything like that?
 one year ago

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

phiBest ResponseYou've already chosen the best response.0
I 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
 one year ago

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

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

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

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

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

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

JonaskBest 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
 one year ago

JonaskBest 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
 one year ago

phiBest ResponseYou've already chosen the best response.0
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
 one year ago

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

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

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