If int frac{1-(cot x)^{2019}}{tan x+(cot x)^{ | vedclass

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(D) Let $I = \int \frac{1-(\cot x)^{2019}}{	an x+(\cot x)^{2020}} dx$. Multiplying numerator and denominator by $\sin x \cos^{2019} x$, we simplify t

Vedclass · Accepted Answer

$\frac{10105}{8}$

Vedclass · Answer

(D) Let $I = \int \frac{1-(\cot x)^{2019}}{	an x+(\cot x)^{2020}} dx$. Multiplying numerator and denominator by $\sin x \cos^{2019} x$, we simplify the integrand. Alternatively, notice that $\frac{d}{dx}(\sin^{2021} x + \cos^{2021} x) = 2021 \sin^{2020} x \cos x - 2021 \cos^{2020} x \sin x = 2021 \sin x \cos x (\sin^{2019} x - \cos^{2019} x)$. Thus, the integral evaluates to $\frac{1}{2021} \ln |\sin^{2021} x + \cos^{2021} x| + c$. Comparing with the given form, $n = 2021$, $f(x) = \sin x$, and $g(x) = \cos x$. We need to evaluate $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}} = 2021 [\sin^4(\frac{\pi}{3}) + \cos^4(\frac{\pi}{3})]$. $= 2021 [(\frac{\sqrt{3}}{2})^4 + (\frac{1}{2})^4] = 2021 [\frac{9}{16} + \frac{1}{16}] = 2021 [\frac{10}{16}] = 2021 [\frac{5}{8}] = \frac{10105}{8}$.

If $\int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx = \frac{1}{n} \ln |(f(x))^n + (g(x))^n| + c$, then the value of $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}}$ is:

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