How can you integrate sec(x^2) with respect to du??

I presume this is supposed to be:

Int[sec^3(x) * tan(x) dx]

= Int[sin(x)/cos^4(x) dx]

Let u = cos(x) ==> du = -sin(x) dx

Int[sec^3(x) * tan(x) dx] = Int[sin(x)/cos^4(x) dx] = Int[1/u^4 * (-du)]

= -Int[1/u^4 du] = -(-1/(3u^3)) + C

= 1/(3*cos^3(x)) + C = (1/3)*sec^3(x) + C

You apparently have the correct answer, but it looks like you had the wrong way to get it.

-Dan