intergration with inverse trig

Printable View

February 4th 2007, 11:07 AM
viet

intergration with inverse trig

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

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

im not able to evaluate it from $\sqrt{2}$ to $2$
February 4th 2007, 11:27 AM
ticbol

Quote:

Originally Posted by viet

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

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

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

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
= pi/12 -----------------answer.
February 4th 2007, 12:01 PM
ThePerfectHacker

Quote:

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
= pi/12 -----------------answer.

It should be,
$\sec^{-1}|x|+C$
Note the absolute value.
In this case it does not matter because of the integration limits are non-negative. But in general it must have an absolute value attached.
February 4th 2007, 09:58 PM
ticbol

Quote:

Originally Posted by ThePerfectHacker

It should be,
$\sec^{-1}|x|+C$
Note the absolute value.
In this case it does not matter because of the integration limits are non-negative. But in general it must have an absolute value attached.

Yeah?

All times are GMT -8. The time now is 12:01 AM.