For I(x)=int frac{sec ^{2} x-2022}{sin ^{2022 | vedclass

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(A) $I(x) = \int \frac{\sec^2 x}{\sin^{2022} x} dx - 2022 \int \frac{1}{\sin^{2022} x} dx$Using integration by parts on the first integral,let $u = \s

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$3^{1010} I \left(\frac{\pi}{3}\right)- I \left(\frac{\pi}{6}\right)=0$

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(A) $I(x) = \int \frac{\sec^2 x}{\sin^{2022} x} dx - 2022 \int \frac{1}{\sin^{2022} x} dx$Using integration by parts on the first integral,let $u = \sin^{-2022} x$ and $dv = \sec^2 x dx$. Then $du = -2022 \sin^{-2023} x \cos x dx$ and $v = 	an x$.$I(x) = 	an x \sin^{-2022} x - \int 	an x (-2022 \sin^{-2023} x \cos x) dx - 2022 \int \sin^{-2022} x dx$$I(x) = 	an x \sin^{-2022} x + 2022 \int \frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sin^{2023} x} dx - 2022 \int \sin^{-2022} x dx$$I(x) = 	an x \sin^{-2022} x + 2022 \int \sin^{-2022} x dx - 2022 \int \sin^{-2022} x dx + C$$I(x) = \frac{	an x}{\sin^{2022} x} + C$Given $I(\pi/4) = 2^{1011}$,we have $\frac{1}{\sin^{2022}(\pi/4)} + C = 2^{1011} \implies (\sqrt{2})^{2022} + C = 2^{1011} \implies 2^{1011} + C = 2^{1011} \implies C = 0$.Thus,$I(x) = \frac{	an x}{\sin^{2022} x}$.$I(\pi/6) = \frac{	an(\pi/6)}{\sin^{2022}(\pi/6)} = \frac{1/\sqrt{3}}{(1/2)^{2022}} = \frac{2^{2022}}{\sqrt{3}}$.$I(\pi/3) = \frac{	an(\pi/3)}{\sin^{2022}(\pi/3)} = \frac{\sqrt{3}}{(\sqrt{3}/2)^{2022}} = \frac{\sqrt{3} \cdot 2^{2022}}{3^{1011}} = \frac{2^{2022}}{3^{1010.5}}$.Comparing $I(\pi/3)$ and $I(\pi/6)$,we find $3^{1010} I(\pi/3) = I(\pi/6)$.

For $I(x)=\int \frac{\sec ^{2} x-2022}{\sin ^{2022} x} d x$,if $I\left(\frac{\pi}{4}\right)=2^{1011}$,then

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