Can you help me solve the integral of sin(x)/[cos^3 (x)] dx in three different ways?

So far I have:a.) tanx sec^2x dx

let u = tanx

du = sec^2x dx

= integral of u du

= (u^2)/2

= 1/2 tan^2 (x) + C

b.) sinx/cos^3x dx

(sinx)(cosx)^-3 dx

let u = cos x

du = -sinx

= - integral of u^-3

= 1/2 (cosx)^-2

= 1/2 sec^2x + C

I can't think of another way of solving it. Thanks in advance!