If int_0^{2024 pi} frac{2023^{sin ^2 x}}{2023 | vedclass

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(B) Let $I = \int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$. Since the period of the integrand is $\pi$,we can writ

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$2025$

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(B) Let $I = \int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$. Since the period of the integrand is $\pi$,we can write $I = 2024 \int_0^{\pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$. Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$,we have $I = 2024 	imes 2 \int_0^{\pi/2} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$. Let $J = \int_0^{\pi/2} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $J = \int_0^{\pi/2} \frac{2023^{\cos ^2 x}}{2023^{\cos ^2 x}+2023^{\sin ^2 x}} d x$. Adding the two expressions for $J$,we get $2J = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$,so $J = \frac{\pi}{4}$. Thus,$I = 2024 	imes 2 	imes \frac{\pi}{4} = 1012 \pi$. Given $k = I = 1012 \pi$,we calculate $\frac{2k}{\pi} + 1 = \frac{2(1012 \pi)}{\pi} + 1 = 2024 + 1 = 2025$.

If $\int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x=k$,then $\left(\frac{2 k}{\pi}+1\right)=$

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