int frac{operatorname{cosec}^2 x-2022}{cos ^{ | vedclass

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Question

(B) Let $I = \int \frac{\operatorname{cosec}^2 x - 2022}{\cos^{2022} x} dx$.We can rewrite the integral as $I = \int \cos^{-2022} x \operatorname{cose

Vedclass · Accepted Answer

$-2^{1011}$

Vedclass · Answer

(B) Let $I = \int \frac{\operatorname{cosec}^2 x - 2022}{\cos^{2022} x} dx$.We can rewrite the integral as $I = \int \cos^{-2022} x \operatorname{cosec}^2 x dx - 2022 \int \sec^{2022} x dx$.Using integration by parts on the first term,let $u = \cos^{-2022} x$ and $dv = \operatorname{cosec}^2 x dx$.Then $du = -2022 \cos^{-2023} x (-\sin x) dx = 2022 \cos^{-2023} x \sin x dx$ and $v = -\cot x$.Applying the formula $\int u dv = uv - \int v du$:$I = \cos^{-2022} x (-\cot x) - \int (-\cot x) (2022 \cos^{-2023} x \sin x) dx - 2022 \int \sec^{2022} x dx$.Since $\cot x \sin x = \cos x$,the integral becomes:$I = -\frac{\cot x}{\cos^{2022} x} + 2022 \int \frac{\cos x}{\cos^{2023} x} dx - 2022 \int \sec^{2022} x dx$.$I = -\frac{\cot x}{\cos^{2022} x} + 2022 \int \sec^{2022} x dx - 2022 \int \sec^{2022} x dx + C$.$I = -\frac{\cot x}{\cos^{2022} x} + C$.Thus,$f(x) = -\frac{\cot x}{\cos^{2022} x}$.Evaluating at $x = \pi/4$: $f(\pi/4) = -\frac{\cot(\pi/4)}{\cos^{2022}(\pi/4)} = -\frac{1}{(1/\sqrt{2})^{2022}} = -\frac{1}{(1/2)^{1011}} = -2^{1011}$.

$\int \frac{\operatorname{cosec}^2 x-2022}{\cos ^{2022} x} d x=f(x)+C \Rightarrow f(\pi / 4)=$

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