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!