integrate 1/( x ( sqrt rt (x2 -1 ) ) ) from √2 to 2

Mar 2014
909
2
malaysia
i gt it = arc sec x , then i sub √2 and 2 inside , i gt an error , why is it wrong ?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
What do you mean you got an error? arcsec(2)- arcsec(√2)= 1.047...- 0.785= 0.738. What did you get?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$\text{arcsec}(2)-\text{arcsec}(\sqrt{2}) = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}$
 
Dec 2013
2,002
757
Colombia
My guess is that it ought to be \(\displaystyle {\pi \over 3} - {\pi \over 4} = {\pi \over 12} = 0.2618 \text{ radians}\)
Did you have your calculator set to degrees?
 
Mar 2014
909
2
malaysia
i pressed 1/ (arc cos x ) on my calc , i gt math error
 
Dec 2013
2,002
757
Colombia
\(\displaystyle \begin{aligned}y = \sec^{-1} x \implies \sec y &= x \\ {1 \over \cos y} &= x \\ \cos y &= {1 \over x} \\ y &= \cos^{-1} {1 \over x} = \sec^{-1} x\end{aligned} \)
 
Mar 2014
909
2
malaysia
\(\displaystyle \begin{aligned}y &= \cos^{-1} {1 \over x} = \sec^{-1} x[/I][/B]\end{aligned} \)
i know that sec x = 1 / cos x .... but how can \(\displaystyle y &= \cos^{-1} {1 \over x} = \sec^{-1} x[\Tex] ??\)
 
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