If int frac{1 - (cot x)^{2021}}{tan x + (cot | vedclass

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(D) Let $I = \int \frac{1 - (\cot x)^{2021}}{	an x + (\cot x)^{2022}} dx$. Multiplying the numerator and denominator by $(\sin x)^{2022} \cos x$,we g

Vedclass · Accepted Answer

$2023$

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(D) Let $I = \int \frac{1 - (\cot x)^{2021}}{	an x + (\cot x)^{2022}} dx$. Multiplying the numerator and denominator by $(\sin x)^{2022} \cos x$,we get: $I = \int \frac{1 - \frac{(\cos x)^{2021}}{(\sin x)^{2021}}}{\frac{\sin x}{\cos x} + \frac{(\cos x)^{2022}}{(\sin x)^{2022}}} dx = \int \frac{(\sin x)^{2021} - (\cos x)^{2021}}{(\sin x)^{2021}} \cdot \frac{(\sin x)^{2022} \cos x}{(\sin x)^{2023} + (\cos x)^{2023}} dx$. Simplifying the expression: $I = \int \frac{(\sin x)^{2021} - (\cos x)^{2021}}{1} \cdot \frac{\sin x \cos x}{(\sin x)^{2023} + (\cos x)^{2023}} dx$. $I = \int \frac{(\sin x)^{2022} \cos x - (\cos x)^{2022} \sin x}{(\sin x)^{2023} + (\cos x)^{2023}} dx$. Let $f(x) = (\sin x)^{2023} + (\cos x)^{2023}$. Then $f'(x) = 2023(\sin x)^{2022} \cos x - 2023(\cos x)^{2022} \sin x = 2023 [(\sin x)^{2022} \cos x - (\cos x)^{2022} \sin x]$. Thus,$I = \frac{1}{2023} \int \frac{f'(x)}{f(x)} dx = \frac{1}{2023} \ln |f(x)| + c$. Comparing this with the given form,we get $A = 2023$.

If $\int \frac{1 - (\cot x)^{2021}}{\tan x + (\cot x)^{2022}} dx = \frac{1}{A} \log |(\sin x)^{2023} + (\cos x)^{2023}| + c$,then $A = . . . . . .$

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