hi can u help me integrate?

Integral x(secx)^2 dx = [xtanx] - Integral[tanx. 1] dx

= [xtanx] - ln[secx] + c

I used integral u dv = [uv] - integral [v du]

int. by parts

hope this helps

OshB: I would increase the exponent by 1 and put the new exponent under X: (X/3)(secX)^3

∫ x(secx)^2 dx

Integration by parts

dv=(sec x)^2

v= tan x

u= x

du=1

∫ u dv = uv - ∫ vdu

x tan x - ∫ tan x (1) dx

x tan x - ∫ tan x dx ----(1)

∫ tan x dx = -ln(cos x) ---(2)

Plug (2) into (1)

x tan x + ln ( cos x) + C