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#.