∫(0→π) √(sin²x - sin⁴x) dx
= ∫(0→π) √[sin²x(1 - sin²x)] dx
= ∫(0→π) √(sin²xcos²x) dx
= ∫(0→π) √[(1/4)(2sinxcosx)²] dx
= (1/2)∫(0→π) √(sin²2x) dx
= (1/2)∫(0→π) |sin2x| dx
= (1/2)[∫(0→π/2) sin2x dx - ∫(π/2→π) sin2x dx]
= (1/4)[cos2x]：(π/2→π) - (1/4)[cos2x]：(0→π/2)
= (1/4)[(1) - (- 1)] - (1/4)[(- 1) - (1)]
= 1