int_{pi / 5}^{3 pi / 10} frac{d x}{sec ^2 x+l | vedclass

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(A) Let $I = \int_{\pi/5}^{3\pi/10} \frac{dx}{\sec^2 x + (	an^{2022} x - 1)(\sec^2 x - 1)}$.Since $\sec^2 x = 1 + 	an^2 x$ and $\sec^2 x - 1 = 	an^

Vedclass · Accepted Answer

$\frac{\pi}{20}$

Vedclass · Answer

(A) Let $I = \int_{\pi/5}^{3\pi/10} \frac{dx}{\sec^2 x + (	an^{2022} x - 1)(\sec^2 x - 1)}$.Since $\sec^2 x = 1 + 	an^2 x$ and $\sec^2 x - 1 = 	an^2 x$,we have:$I = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + 	an^2 x + (	an^{2022} x - 1)	an^2 x}$$I = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + 	an^2 x + 	an^{2024} x - 	an^2 x} = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + 	an^{2024} x}$ ... $(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a = \pi/5$ and $b = 3\pi/10$,$a+b = \pi/2$:$I = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + 	an^{2024}(\pi/2 - x)} = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + \cot^{2024} x}$$I = \int_{\pi/5}^{3\pi/10} \frac{	an^{2024} x}{1 + 	an^{2024} x} dx$ ... (ii)Adding $(i)$ and (ii):$2I = \int_{\pi/5}^{3\pi/10} \frac{1 + 	an^{2024} x}{1 + 	an^{2024} x} dx = \int_{\pi/5}^{3\pi/10} dx = \frac{3\pi}{10} - \frac{\pi}{5} = \frac{\pi}{10}$Therefore,$I = \frac{\pi}{20}$.

$\int_{\pi / 5}^{3 \pi / 10} \frac{d x}{\sec ^2 x+\left(\tan ^{2022} x-1\right)\left(\sec ^2 x-1\right)}=$

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