Kerala KEAM 5 June 2024 Solved Paper

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).Quick Tip: 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.