int_0^(π/2)sin^2x dx 

আমরা \( \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \) এর মান নির্ণয় করব।

আমরা জানি, \( \cos 2x = 1 - 2\sin^2 x \)।

সুতরাং, \( \sin^2 x = \frac{1 - \cos 2x}{2} \).

অতএব, \( \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx \)

\( = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx \)

\( = \frac{1}{2} \left[ \int_{0}^{\frac{\pi}{2}} 1 \, dx - \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx \right] \)

\( = \frac{1}{2} \left[ x \Big|_0^{\frac{\pi}{2}} - \frac{\sin 2x}{2} \Big|_0^{\frac{\pi}{2}} \right] \)

\( = \frac{1}{2} \left[ \frac{\pi}{2} - 0 - \frac{1}{2} (\sin \pi - \sin 0) \right] \)

\( = \frac{1}{2} \left[ \frac{\pi}{2} - \frac{1}{2} (0 - 0) \right] \)

\( = \frac{1}{2} \cdot \frac{\pi}{2} \)

\( = \frac{\pi}{4} \) 🎉

সুতরাং, \( \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{4} \) 🥳।