∫x.√(sinx^2-sinx^4) dx (下限0 上限π)
2个回答

∫(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