UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$ and a trigonometric substitution to evaluate the integral $I_1$, then compute $8I_1^2$.Explanation:Step 1: Write the given integral: $I_1 = \int_{0}^{\pi} \frac{x}{1 + \cos^2 x} \, dx$.Step 2: Apply the property: $I_1 = \int_{0}^{\pi} \frac{\pi - x}{1 + \cos^2(\pi - x)} \, dx = \int_{0}^{\pi} \frac{\pi - x}{1 + \cos^2 x} \, dx$.Step 3: Add the two expressions for $I_1$: $2I_1 = \int_{0}^{\pi} \frac{\pi}{1 + \cos^2 x} \, dx$.Step 4: Use symmetry about $x = \pi/2$: $2I_1 = 2 \int_{0}^{\pi/2} \frac{\pi}{1 + \cos^2 x} \, dx$, so $I_1 = \pi \int_{0}^{\pi/2} \frac{1}{1 + \cos^2 x} \, dx$.Step 5: Rewrite integrand using $\sec^2 x$: $\frac{1}{1 + \cos^2 x} = \frac{\sec^2 x}{1 + \sec^2 x} = \frac{\sec^2 x}{2 + \tan^2 x}$. Thus $I_1 = \pi \int_{0}^{\pi/2} \frac{\sec^2 x}{2 + \tan^2 x} \, dx$.Step 6: Substitute $t = \tan x$, $dt = \sec^2 x \, dx$. Limits: $x=0 \to t=0$, $x=\pi/2 \to t=\infty$. Then $I_1 = \pi \int_{0}^{\infty} \frac{dt}{2 + t^2}$.Step 7: Evaluate: $\int \frac{dt}{2+t^2} = \frac{1}{\sqrt{2}} \tan^{-1} \frac{t}{\sqrt{2}}$. So $I_1 = \frac{\pi}{\sqrt{2}} \left[ \tan^{-1} \frac{t}{\sqrt{2}} \right]_{0}^{\infty} = \frac{\pi}{\sqrt{2}} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi^2}{2\sqrt{2}}$.Step 8: Compute $8I_1^2$: $8I_1^2 = 8 \times \left( \frac{\pi^2}{2\sqrt{2}} \right)^2 = 8 \times \frac{\pi^4}{8} = \pi^4$.Answer:$\pi^4$; corresponds to option D.