Question

Question #abe90

Answer 1

`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 further`int 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.