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

#### xl5899

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

#### HallsofIvy

MHF Helper
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
$\text{arcsec}(2)-\text{arcsec}(\sqrt{2}) = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}$

#### Archie

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?

#### xl5899

i pressed 1/ (arc cos x ) on my calc , i gt math error

#### skeeter

MHF Helper
i pressed 1/ (arc cos x ) on my calc , i gt math error
$\text{arcsec}(x) = \arccos\left(\dfrac{1}{x}\right)$

#### Archie

i pressed 1/ (arc cos x ) on my calc , i gt math error
This is why it's better to know the principal values of $\sin$ and $\cos$.

#### xl5899

$\text{arcsec}(x) = \arccos\left(\dfrac{1}{x}\right)$
.why it shouldnt become like this ?

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#### Archie

\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}

#### xl5899

\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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