Question #abe90

1 Answer
A.Raz
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 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.

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