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[jeca86]

Use the substitution x=sectheta, where 0<theta<pi/2 or pi<theta<3pi/2, to evaluate

the integral of the (square root(x^2 -1))/x^4 *dx

i have:

x=sectheta
dx=sectheta*tantheta*dtheta

integral of (square root((sec^2theta)-1)/sec^4theta)*(sectheta*tantheta*dtheta)

=(tantheta/sec^3theta)*(tantheta*dtheta)

=(tan^2theta/sec^3theta)*dtheta

=(sec^2theta-1)/(sectheta(1+tan^2theta))*dtheta

this is where i get lost.[/img]

 

[Unco]

Use the substitution x=sectheta, where 0<theta<pi/2 or pi<theta<3pi/2, to evaluate

the integral of the (square root(x^2 -1))/x^4 *dx

i have:

x=sectheta
dx=sectheta*tantheta*dtheta

integral of (square root((sec^2theta)-1)/sec^4theta)*(sectheta*tantheta*dtheta)

=(tantheta/sec^3theta)*(tantheta*dtheta)

=(tan^2theta/sec^3theta)*dtheta (*)

=(sec^2theta-1)/(sectheta(1+tan^2theta))*dtheta

this is where i get lost.[/img]

(*) You could simpify this to\(\displaystyle \mbox{ \int \sin^2{\theta} \cos{\theta} d\theta}\).

(Then think u-substitution.)