Indefinite Integral (u sub)

[keega]

I'm a little stuck on this integral:

Given problem (obviously the integral symbol precedes the "sec":

sec^2x[sqrt]tan(x)[/sqrt]dx

and here's what I've done:

let u = sec(x)
let du = tan(x)dx

=[int]u^2 * du dx
=u^3/3 * du dx
=sec^3(x)/3 + c

Is this the right answer?

Thanks in advance!

 

[pka]

Let u=tan(x) then du=sec²(x)(x)dx.
You get √(u)du

 

[keega]

okies, I can take it from there...thanks much!