Integrate It!

$\begin{aligned}\text{ Let, }I&=\int \dfrac{1+\cos8x}{\cot2x-\tan2x} dx\\ &=\int \dfrac{2\cos^24x}{\dfrac{\cos2x}{\sin2x}-\dfrac{\sin2x}{\cos2x}}dx\\ &=\int \dfrac{\cos^24x}{\dfrac{\cos^22x-\sin^22x}{2\cdot\sin2x\cdot\cos2x}}dx\\ &=\int \dfrac{\cos^24x}{\left(\dfrac{\cos4x}{\sin4x}\right)}dx\\ &=\int\sin4x\cdot\cos4x\hspace{2mm} dx \hspace{10mm}\color{#3D99F6}\text{Let,}\sin4x=t ,dt=4\cos4x\hspace{2mm} dx\\ &=\dfrac{1}{4}\int tdt\\ \implies I&=\dfrac{t^2}{8}+C_{1}=\dfrac{\sin^24x}{8}+C_{1}\\ \\ &\text{Manipulating the constant }C_{1},\text{we can rewrite I as,}\\ \\ I&=\dfrac{2\sin^24x-1}{16}+C_{2} \hspace{10mm} \color{#3D99F6}\text{Where,} C_{2}=C_{1}+\dfrac{1}{16}\\ &=\dfrac{-1}{16}\cos8x+C_{2}\\ \implies&\color{#D61F06}\boxed{x=-1},\boxed{y=16}\\ |x+y|&=|-1+16|=\color{#D61F06}\boxed{15}\end{aligned}$