Something needs to be reduced

$\large \displaystyle 2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{2017}(x) dx$

$\color{#20A900} \large \displaystyle \cot(x) = u \implies \sin(x) = \frac{1}{\sqrt{u^2+1}}$

$\color{#20A900} \large \displaystyle - \csc^2(x) dx = du \implies dx = - \sin^2(x) du \implies dx = - \frac{1}{u^2+1} du$

$\large \displaystyle = -2 \int_1^0 \frac{u^{2017}}{u^2+1} du$

$\large \displaystyle = 2 \int_0^1 \frac{u^{2017}}{u^2+1} du$

$\large \displaystyle = 2 \int_0^1 \frac{u^{2017} + u^{2015} - u^{2015} - u^{2013} + ... - u^3 - u + u}{u^2+1} du$

$\large \displaystyle = 2 \int_0^1\left [ \frac{u^{2017} + u^{2015}}{u^2+1} - \frac{u^{2015} + u^{2013}}{u^2+1} + ... - \frac{u^3 + u}{u^2+1} + \frac{u}{u^2+1} \right ] du$

$\large \displaystyle = 2 \int_0^1\left [ u^{2015} - u^{2013} + u^{2011} - ... - u + \frac{u}{u^2+1} \right ] du$

$\large \displaystyle = 2 \left [ \frac{1}{2016} - \frac{1}{2014} + \frac{1}{2012} - ... - \frac{1}{2} \right ] + \int_0^1 \frac{2u}{u^2+1} du$

$\large \displaystyle = \sum_{n=1}^{1008} \frac{(-1)^n}{n} + \ln(u^2+1) \Bigg| _0^1$

$\color{#20A900} \boxed{ \large \displaystyle = \sum_{n=1}^{1008} \frac{(-1)^n}{n} + \ln(2) }$

Then:

$\color{#3D99F6} \large \displaystyle a = 1008, b = 2, \boxed{\large \displaystyle a+b=1010}$

The integral can be solved using the reduction formula below:

$\begin{aligned} \int_\frac \pi 4^\frac \pi 2\cot^k x \ dx & = - \frac {\cot^{k-1} x}{k-1} \bigg|_\frac \pi 4^\frac \pi 2 - \int _\frac \pi 4^\frac \pi 2 \cot^{k-2} x \ dx = \frac 1{k-1} - \int _\frac \pi 4^\frac \pi 2 \cot^{k-2} x \ dx \end{aligned}$

Let $\displaystyle I_k = 2 \int_\frac \pi 4^\frac \pi 2\cot^k x \ dx$ ; then we have $\displaystyle I_k = \frac 2{k-1} - I_{k-2}$ .

$\begin{aligned} I_1 & = 2 \int_\frac \pi 4^\frac \pi 2\cot x \ dx = \ln 2 \\ I_3 & = (-1)(-1 + \ln 2) \\ I_5 & = \left(-1+ \frac 12 + \ln 2\right) \\ I_7 & =(-1)\left(-1+ \frac 12 - \frac 13 + \ln 2 \right) \\ \cdots & = \cdots \\ \implies I_{2a+1} & =(-1)^a \left( \sum_{n=1}^a \frac {(-1)^n}n + \ln 2 \right) \\ \implies I_{2017} & =(-1)^{\color{#3D99F6}1008} \left( \sum_{n=1}^{\color{#3D99F6}1008} \frac {(-1)^n}n + \ln {\color{#3D99F6}2} \right) \end{aligned}$

$\implies a + b = 1008 + 2 = \boxed{1010}$