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If $\int_{}^{} \frac{\sec^{2} x - 2023}{\sin^{2023} x} dx = \frac{P (x)}{{(sinx)}^{2023}} + C,$ then which of the following is true? (‘C’ is integration constant and P(0) = 0 )

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$P (\frac{π}{2}) = 1$

$P (\frac{π}{4}) = \sqrt{2}$

$P (\frac{π}{3}) = \sqrt{3}$

$P (\frac{π}{8}) + P (\frac{π}{4}) = \sqrt{2} + 1$

answer is B.

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Detailed Solution

$\int \frac{s e c^{2} x}{\sin^{2023} x} d x - \int \frac{2023}{\sin^{2023} x} d x$

$= \frac{\tan x}{\sin^{2023} x} + 2023 \int \frac{\tan x \cos x}{\sin^{2024} x} d x - \int \frac{2023}{\sin^{2023} x} d x$
$= \frac{tanx}{{(sinx)}^{2023}} + C$

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