Question

The value of \[ \int_0^{\frac{\pi}{2}} \frac{\cos^{2024} x}{\sin^{2024} x + \cos^{2024} x} \, dx \] is equal to:

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( 2\pi \)
• D. \( \pi \)
• E. \( \frac{\pi}{3} \)

Hint:

When evaluating integrals with symmetric bounds and functions, use substitution to exploit symmetry. In this case, substituting \( x \to \frac{\pi}{2} - x \) allows the integral to simplify.

Answer 1

The Correct Option is A

We are asked to solve the integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{\cos^{2024} x}{\sin^{2024} x + \cos^{2024} x} \, dx \] Step 1: Notice that the integrand has symmetry. Let's perform a substitution \( x \to \frac{\pi}{2} - x \). 
Under this substitution:
\[ \sin \left( \frac{\pi}{2} - x \right) = \cos x, \quad \cos \left( \frac{\pi}{2} - x \right) = \sin x. \] Step 2: Substituting in the integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{\cos^{2024} \left( \frac{\pi}{2} - x \right)}{\sin^{2024} \left( \frac{\pi}{2} - x \right) + \cos^{2024} \left( \frac{\pi}{2} - x \right)} \, dx \] This transforms the integral to: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^{2024} x}{\cos^{2024} x + \sin^{2024} x} \, dx. \] Step 3: Adding the two equations, we get: \[ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{\cos^{2024} x}{\sin^{2024} x + \cos^{2024} x} + \frac{\sin^{2024} x}{\sin^{2024} x + \cos^{2024} x} \right) \, dx \] Simplifying the integrand: \[ 2I = \int_0^{\frac{\pi}{2}} 1 \, dx \] \[ 2I = \frac{\pi}{2} \] Step 4: Therefore, \( I = \frac{\pi}{4} \).
Thus, the value of the integral is \( \frac{\pi}{4} \).
Therefore, the correct answer is option (A).