Explanation: 
use π/2 instead of π.
Another Explanation (5):
সমাধান: ∫0π/2 sin2x dx = ?
আমরা জানি, cos 2x = 1 - 2sin2x
সুতরাং, sin2x = (1 - cos 2x)/2
এখন,
∫0π/2 sin2x dx = ∫0π/2 (1 - cos 2x)/2 dx
= (1/2) ∫0π/2 (1 - cos 2x) dx
= (1/2) [∫0π/2 1 dx - ∫0π/2 cos 2x dx]
= (1/2) [x |0π/2 - (sin 2x)/2 |0π/2]
= (1/2) [(π/2 - 0) - (sin π - sin 0)/2]
= (1/2) [π/2 - (0 - 0)/2]
= (1/2) [π/2 - 0]
= π/4
অতএব, ∫0π/2 sin2x dx = π/4 🥳