If ∫ 1 - ( cot x ) 2021 tan x + ( cot x ) 2022 d x = 1 A log ( sin x ) 2023 + ( cos x ) 2023 + c ,…

Given that 

$\int \frac{1-{\left(cotx\right)}^{2021}}{\tan x +{\left(cotx\right)}^{2022} }dx$

$=\int \frac{{\left(\sin x\right)}^{2021} - {\left(\cos x\right)}^{2021}}{{\left(\sin x\right)}^{2021}}\times \frac{{\left(\sin x\right)}^{2022}\cos x}{{\left(\sin x\right)}^{2023}+\cos x^{2023}}dx$

$=\int \frac{{\left(\sin x\right)}^{2021} - {\left(\cos x\right)}^{2021}}{1}\times \frac{\sin x\cos x}{{\left(\sin x\right)}^{2023}+\cos x^{2023}}dx$

$=\int \frac{{\left(\sin x\right)}^{2022}\cos x-{\left(\cos x\right)}^{2022}\sin x}{{\left(\sin x\right)}^{2023}+{\left(\cos x\right)}^{2023}}dx$

$= \frac{1}{2023}\int \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}dx = \frac{1}{2023}\ln \left(f\left(x\right)\right)+ c$

So $A = 2023.$