Question

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

• A. \( 2023 \)
• B. \( 2025 \)
• C. \( 2022 \)
• D. \( 2024 \)

Hint:

Use the property \( \int_0^a f(x) dx = \int_0^a f(a - x) dx \) to simplify the integral. Also, use the periodicity of the trigonometric functions to reduce the limits of integration. The identity \( \sin^2 x + \cos^2 x = 1 \) is helpful.

Answer 1

The Correct Option is B

Let \( I = \int_0^{2024\pi} \frac{2023^{\sin^2 x}}{2023^{\sin^2 x} + 2023^{\cos^2 x}} dx \).
We use the property \( \int_0^{na} f(x) dx = n \int_0^a f(x) dx \) if \( f(x + a) = f(x) \).
The period of \( \sin^2 x \) and \( \cos^2 x \) is \( \pi \).
So, \( I = 2024 \int_0^{\pi} \frac{2023^{\sin^2 x}}{2023^{\sin^2 x} + 2023^{\cos^2 x}} dx \).
Let \( J = \int_0^{\pi} \frac{2023^{\sin^2 x}}{2023^{\sin^2 x} + 2023^{\cos^2 x}} dx \).
Using the property \( \int_0^a f(x) dx = \int_0^a f(a - x) dx \): $$ J = \int_0^{\pi} \frac{2023^{\sin^2 (\pi - x)}}{2023^{\sin^2 (\pi - x)} + 2023^{\cos^2 (\pi - x)}} dx = \int_0^{\pi} \frac{2023^{\sin^2 x}}{2023^{\sin^2 x} + 2023^{(-\cos x)^2}} dx $$ $$ J = \int_0^{\pi} \frac{2023^{\sin^2 x}}{2023^{\sin^2 x} + 2023^{\cos^2 x}} dx $$ This doesn't give new information.
Let's use \( \int_0^a f(x) dx = \int_0^a f(a - x) dx \) with \( a = \pi \): $$ J = \int_0^{\pi} \frac{2023^{\sin^2 (\pi - x)}}{2023^{\sin^2 (\pi - x)} + 2023^{\cos^2 (\pi - x)}} dx = \int_0^{\pi} \frac{2023^{\sin^2 x}}{2023^{\sin^2 x} + 2023^{\cos^2 x}} dx $$ Adding the two expressions for \( J \): $$ 2J = \int_0^{\pi} \frac{2023^{\sin^2 x} + 2023^{\cos^2 x}}{2023^{\sin^2 x} + 2023^{\cos^2 x}} dx = \int_0^{\pi} 1 dx = [x]_0^{\pi} = \pi $$ So, \( 2J = \pi \implies J = \frac{\pi}{2} \).
Now, \( k = 2024 J = 2024 \cdot \frac{\pi}{2} = 1012\pi \).
We need to find \( \left( \frac{2k}{\pi} + 1 \right) \).
$$ \frac{2k}{\pi} + 1 = \frac{2(1012\pi)}{\pi} + 1 = 2 \cdot 1012 + 1 = 2024 + 1 = 2025 $$