How do you find int (1-tanx)^2dx?

1 Answer
May 7, 2018

int ( 1 - tan x )^2\ d x = tan x + 2 ln | cos x | + C

Explanation:

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.