UPSC NDA II 2024 Mathematics Paper

Concept:Use the symmetry property of definite integrals: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.This simplifies integrals with trigonometric expressions over symmetric limits.Also use the known standard result: $\int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx = \ln 2$.Explanation:Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x + 1} dx$.Apply the substitution $x \to \frac{\pi}{2} - x$.Then $I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x + 1} dx$.Add the two expressions: $2I = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x + 1} dx$.Rewrite the numerator: $\sin x + \cos x = (\sin x + \cos x + 1) - 1$.Thus $2I = \int_0^{\frac{\pi}{2}} \left(1 - \frac{1}{\sin x + \cos x + 1}\right) dx$.Separate the integrals: $2I = \int_0^{\frac{\pi}{2}} 1\, dx - \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx$.The first integral equals $\frac{\pi}{2}$.The second integral equals $\ln 2$ (standard result).Hence $2I = \frac{\pi}{2} - \ln 2$.Divide by 2: $I = \frac{\pi}{4} - \frac{\ln 2}{2}$.Answer:$\frac{\pi}{4} - \frac{\ln 2}{2}$ (Option C).