Differential Equation Problem: $2\cot x\frac{\mathrm{d}y}{\mathrm{d}x}=4-y^2$ Solve in terms of $\sec^2 x $$$2\cot x\frac{\mathrm{d}y}{\mathrm{d}x}=4-y^2$$
if $y=0$ when $x=\pi/3$

My attempt $$\int\tan x\,\mathrm{d}x=\int\frac{2}{4-y^2}\,\mathrm{d}y$$ 
Then by using partial fractions on the RHS 
\begin{align}\ln\vert\sec x\vert+C_0&=\frac{1}{2}\left(\int\frac{1}{2-y}+\frac{1}{2+y}\,\mathrm{d}y\right)\\ &=\frac{1}{2}\left(\ln\vert2-y\vert+\ln\vert2+y\vert\right)\\&=\frac{1}{2}\ln\vert4-y^2\vert\end{align}
By taking exponents of both sides
$$\vert\sec^2x\vert+C_1=\vert4-y^2\vert$$
$$\sec^2(\pi/3)+C_1=4\rightarrow C_1=0$$ and the solution is $\sec^2x=\vert4-y^2\vert$
However the actual answer is $\sec^2x=\frac{1}{2}\ln\left\vert\frac{2+y}{2-y}\right\vert+8$

My question is have I gone wrong and if so where? I have tried using WolframAlpha though it does not seem to be very helpful with problems like this. My only thoughts are that maybe rearranging altered the equation since after $y\neq2.$
 A: $$\int\tan x\,\mathrm{d}x=\int\frac{2}{4-y^2}\,\mathrm{d}y$$ 
$$\int\tan x\,\mathrm{d}x=\int\frac{\sin x}{\cos x}dx=-\int \frac{d(\cos x)}{\cos x}=-\ln x+C=\ln C_0-\ln \cos x=\ln \frac{C_0}{\cos x}$$
$$2\int \frac{dy}{4-y^2}=\frac12\int\left(\frac1{2-y}+\frac1{2+y}\right)dy=\frac12(\ln|2+y|-\ln|2-y|)=\frac12\ln|\frac{2+y}{2-y}|=$$
$$=\ln \sqrt{\frac{2+y}{2-y}}$$
Then $$\ln \frac{C_0}{\cos x}=\ln \sqrt{\frac{2+y}{2-y}}$$
$$\frac{C_0}{\cos x}= \sqrt{\frac{2+y}{2-y}}$$
A: Note:$$\int \frac{dy}{2-y}=-\ln|2-y|.$$ That's where you gone wrong.
A: Your error comes from the integral of the first partial fraction:
$$\int\frac{\mathrm d\mkern1mu y}{2-y}=-\ln\lvert 2-y\rvert,\enspace\text{so that eventually}\enspace \int\frac{2\mathrm{d}\mkern1mu y}{4-y^2}=\frac12\bigl(\ln\lvert 2+y\rvert-\ln\lvert 2-y\rvert\bigr).$$
That said, everyone should know (without having to reinvent the wheel) that
$$\int\frac{\mathrm{d}\mkern1mu x}{a^2-x^2}=\frac1{2a}\ln\biggl\lvert\frac{a+x}{a-x}\biggr\rvert=\frac1a\operatorname{artanh}\Bigl(\frac xa\Bigr).$$
