Unseen integral?

Consider the following integral:

$\begin{aligned} I_n & = \int_0^\frac \pi 4 \frac {\ln(\cot x)\sin^n 2x}{\left(\sin^{n+1}x+\cos^{n+1}x \right)^2}\ dx \\ & = \int_0^\frac \pi 4 \frac {-\ln(\tan x)2^n\sin^n x\cos^n x}{\left(\sin^{n+1}x+\cos^{n+1}x \right)^2}\ dx & \small \color{#3D99F6} \text{Divide up and down by }\cos^ {2n+2}x \\ & = \int_0^\frac \pi 4 \frac {-2^n\ln(\tan x)\tan^n x \sec^2 x}{\left(\tan^{n+1}x+1 \right)^2}\ dx & \small \color{#3D99F6} \text{Let }t = \tan x \implies dt = \sec^2 x \ dx \\ & = \int_0^1 \frac {-2^n t^n \ln t}{\left(t^{n+1}+1 \right)^2}\ dt & \small \color{#3D99F6} \text{By integration by parts} \\ & = \frac {2^n \ln t}{(n+1)\left(t^{n+1}+1 \right)} \bigg|_0^1 - \frac {2^n}{n+1} \int_0^1 \frac {dt}{t\left(t^{n+1}+1 \right)} \\ & = \frac {2^n \ln t}{(n+1)\left(t^{n+1}+1 \right)} \bigg|_0^1 - \frac {2^n}{n+1} \color{#3D99F6} \int_0^1 \frac {dt}{t^{n+2}\left(1+\frac 1{t^{n+1}} \right)} & \small \color{#3D99F6} \text{Let }u = 1+\frac 1{t^{n+1}} \implies du = - \frac {n+1}{t^{n+2}}dt \\ & = \frac {2^n \ln t}{(n+1)\left(t^{n+1}+1 \right)} \bigg|_0^1 \color{#3D99F6} + \frac {2^n}{(n+1)^2} \int_\infty^2 \frac {du}u \\ & = \frac {2^n \ln t}{(n+1)\left(t^{n+1}+1 \right)} \bigg|_0^1 + \frac {2^n\ln u}{(n+1)^2} \bigg|_\infty^2 \\ & = \frac {2^n \ln t}{(n+1)\left(t^{n+1}+1 \right)} + \frac {2^n\ln \left(1+\frac 1{t^{n+1}}\right)}{(n+1)^2} \bigg|_0^1 \\ & = \frac {2^n}{n+1} \left[ \frac {\ln t}{t^{n+1}+1} + \frac {\ln \left(t^{n+1}+1\right)}{n+1} - \ln t \right]_0^1 \\ & = \frac {2^n \ln 2}{(n+1)^2} \end{aligned}$

$\implies I_{2008} = \dfrac {2^{2008}\ln 2}{2009^2} \implies a +b+c - 4000 = 2+2008+2009-4000 = \boxed{19}$