Question #abe90

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Question #abe90

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Answer 1 · 2017-04-09T17:48:18

$\frac{1}{tan a} [ln | sec (x + a) | - ln | sec x |]$ $1/tan a[ ln abs(sec (x+a)) - ln abs secx]$

Explanation:

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

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

$(1 + tan (x + a) tan x) = \frac{tan (x + a) - (tan x)}{tan a}$ $(1+tan( x+a) tanx)= [tan (x+a) -(tan x)]/tan a$

$\int (1 + tan (x + a) tan x) = \int \frac{tan (x + a) - (tan x)}{tan a}$ $int (1+tan( x+a) tanx)= int[tan (x+a) -(tan x)]/tan a$

$\frac{1}{tan a}$ $1/tan a$ is a constant that can come out of the integral

$\frac{1}{tan a} \int tan (x + a) d x$ $1/tana int tan(x+a)dx$ - $\frac{1}{tan a} \int tan x d x$ $1/tan aint tan x dx$
We use this rule to simplify this further $\int tan x d x = ln | sec x |$ $int tan x dx= ln abssecx$

$\frac{1}{tan a} [ln | sec (x + a) | - ln | sec x |]$ $1/tan a[ ln abs(sec (x+a)) - ln abs secx]$

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

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Question #abe90

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