1. intergration with inverse trig

Evaluate: $\displaystyle \int_{\sqrt{2}}^{2} \dfrac{dx}{x\sqrt{x^{2}-1}}$

$\displaystyle \dfrac{1}{x\sqrt{x^{2}-1}}dx$ is the same as $\displaystyle arcsec(x)$

im not able to evaluate it from $\displaystyle \sqrt{2}$ to $\displaystyle 2$

2. Originally Posted by viet
Evaluate: $\displaystyle \int_{\sqrt{2}}^{2} \dfrac{dx}{x\sqrt{x^{2}-1}}$

$\displaystyle \dfrac{1}{x\sqrt{x^{2}-1}}dx$ is the same as $\displaystyle arcsec(x)$

im not able to evaluate it from $\displaystyle \sqrt{2}$ to $\displaystyle 2$
1 / [x*sqrt(x^2 -1)] dx is not the same as arcsec(x).
Rather, arcsec(x) +C = INT.[1 / x*sqrt(x^2 +1)]dx.

So, after integration,
= arcsec(x) | (sqrt(2) to 2)
= arcsec(2) -arcsec(sqrt(2))
= pi/3 -pi/4

3. Originally Posted by ticbol
1 / [x*sqrt(x^2 -1)] dx is not the same as arcsec(x).
Rather, arcsec(x) +C = INT.[1 / x*sqrt(x^2 +1)]dx.

So, after integration,
= arcsec(x) | (sqrt(2) to 2)
= arcsec(2) -arcsec(sqrt(2))
= pi/3 -pi/4
$\displaystyle \sec^{-1}|x|+C$
$\displaystyle \sec^{-1}|x|+C$