evaluate the indefinite interval:

Question

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

Answer 1

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)