How is $2(\ln \tan x + c)$ simplified to $A \tan^2 x$ where $A =2c$ I'm trying to follow this reasoning:
$1/2 \ln(4 +y^2) = \ln(\tan x) + C$
$\ln(4 +y^2) = 2\ln(\tan x) + \ln A$  ( constant 2C = A)
$4 + y^2 = A \tan^2 x$ 

$\frac{1}{2}ln(4+ y^2)= ln(tan(x))+ C$
Multiply on both sides by 2:
$ln(4+ y^2)= 2ln(tan(x))+ 2C$
Taking $A= e^{2C}$ (NOT A= 2C) we have $2C= ln(A)$ so
$ln(4+ y^2)= ln(tan^2(x))+ ln(A)= ln(Atan^2(x))$
Now take the exponential of both sides:
$e^{ln(4+ y^2)}= e^{ln(Atan^2(x))}$
$4+ y^2= Atan^2(x)$
I have used:
1)  $2ln(x)= ln(x^2)$
2)  $ln(a)+ ln(b)= ln(ab)$
3)  $e^{ln(x)}= x$
