First, we expand the binomial, and make use of the linearity of the integral:
int ( 1 - tan x )^2\ d x = int \ d x - 2 int tan x\ d x + int tan^2 x\ d x =
= x - 2 int tan x\ d x + int tan^2 x\ d x.
Now we have two simple integrals to solve. The first one:
int tan x\ d x =
= int \frac{sin x}{cos x}\ d x =
= [ - int \frac{1}{t}\ d t ]_{t = cos x} =
= - ln | cos x | + C.
And the second one:
int tan^2 x\ d x =
= int \frac{sin^2 x}{cos^2 x}\ d x =
= int \frac{1 - cos^2 x}{cos^2 x}\ d x =
= int \frac{1}{cos^2 x} - 1\ d x =
= tan x - x + C.
Putting it all together, we arrive at
x - 2 int tan x\ d x + int tan^2 x\ d x =
x + 2 ln | cos x | + tan x - x + C=
2 ln | cos x | + tan x + C.