Q104: The value of $\int_0^{\frac{\pi}{2}} \sin^2\frac{x}{2} \cos^2\frac{x}{2} \, dx$ is given... - Mathematics - ICAR AIEEA PG Statistical Science 14 July 2023 Shift-1 Paper | Agrigyan

Thus, $\sin^2\frac{x}{2} \cos^2\frac{x}{2} = \frac{1}{4}\sin^2 x$. The integral becomes: $I = \int_0^{\pi/2} \frac{1}{4}\sin^2 x \, dx = \frac{1}{4} \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx$ $= \frac{1}{8} \left[ x - \frac{\sin 2x}{2} \right]_0^{\pi/2} = \frac{1}{8} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{16}$. Option breakdown:

• $\frac{\pi}{2}$ misses the $1/8$ factor.
• $\frac{\pi}{4}$ misses the $1/4$ factor.
• $\frac{\pi}{16}$ is the exact evaluated value.
• $\pi$ is incorrect.