Show that integral (1 - 1/(x^2))^(1/2)*dx = integral tan^2u du using the substitution x = secu

Ahh scary :S thanks in advance

if x = sec u (sec is 1/cos), then

dx = d(1/cosu) = du sinu/cos^2u

and (1 - 1/(x^2))^(1/2) = (1 - 1/((1/cosu))^2))^(1/2) = (sin^2u)^(1/2) = sin u

thus, (1 - 1/(x^2))^(1/2)dx = sin^2 u/cos^2u du = tan ^2 du