Thread: intergration with inverse trig

Evaluate:

is the same as

im not able to evaluate it from to

Originally Posted by viet

Evaluate:

is the same as

im not able to evaluate it from to

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.

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,

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.

Originally Posted by ThePerfectHacker

It should be,

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?