UPSC NDA II 2023 Mathematics Solved Paper

Concept:Use algebraic manipulation and trigonometric identities to simplify the integrand $\sin^4 x + \cos^4 x$ into a form that is easy to integrate over $0$ to $\pi$.Explanation:First, add and subtract $2\sin^2 x \cos^2 x$ inside the integral:$\int_0^\pi (\sin^4 x + \cos^4 x) dx = \int_0^\pi \big[(\sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x) - 2\sin^2 x \cos^2 x\big] dx$.The expression inside brackets factors as $(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$.Since $\sin^2 x + \cos^2 x = 1$, this becomes $1 - 2\sin^2 x \cos^2 x$.Also, $2\sin x \cos x = \sin 2x$, so $2\sin^2 x \cos^2 x = \frac{1}{2} \sin^2 2x$.Thus the integral simplifies to $\int_0^\pi \left(1 - \frac{1}{2}\sin^2 2x\right) dx$.Now use the identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$ to get:$\int_0^\pi \left[1 - \frac{1}{2} \cdot \frac{1 - \cos 4x}{2}\right] dx = \int_0^\pi \left(1 - \frac{1}{4} + \frac{\cos 4x}{4}\right) dx = \int_0^\pi \left(\frac{3}{4} + \frac{\cos 4x}{4}\right) dx$.Integrate term by term:$\int_0^\pi \frac{3}{4} dx = \frac{3}{4} \cdot \pi = \frac{3\pi}{4}$,and $\int_0^\pi \frac{\cos 4x}{4} dx = \frac{1}{4} \cdot \frac{\sin 4x}{4} \Big|_0^\pi = \frac{1}{16} (\sin 4\pi - \sin 0) = 0$.Therefore, the total value is $\frac{3\pi}{4}$.Answer:$\frac{3\pi}{4}$ (Option B).