evaluate the indefinite interval:
x sec^(2)(9x) dx
i know u=x du=dx dv=sec^(2)(9x) and v = 1/9tan9x....but i don't know where to go from there
much help is appreciated
looks very similar to the other one we did.
let u = x
du/dx = 1
du = dx
let dv = sec^2 (9x) dx
v = (1/9(tan(9x))
so we have
uv - [intg]v du
= (x/9)tan(9x) + (1/9)ln(cosx)