Question #abe90

1 Answer
Apr 9, 2017

1/tan a[ ln abs(sec (x+a)) - ln abs secx]

Explanation:

Using the rule: tan(A-B)= (tan A -tan B)/(1+tanA tanB)
Set:
A=x+a
B=x
Substitute these values into the above rule: tan(a)= [tan (x+a) -(tan x)]/(1+tan( x+a) tanx)

The denominator is the antiderivative we want to find.
Note tan a is a constant

(1+tan( x+a) tanx)= [tan (x+a) -(tan x)]/tan a

int (1+tan( x+a) tanx)= int[tan (x+a) -(tan x)]/tan a

1/tan a is a constant that can come out of the integral

1/tana int tan(x+a)dx - 1/tan aint tan x dx
We use this rule to simplify this furtherint tan x dx= ln abssecx

1/tan a[ ln abs(sec (x+a)) - ln abs secx]

I did not assume that you meant int(1+tanx)(tan a +x) which would result in a different answer.