求 ∫(上限π/2,下限0 )√(1-sin2x) dx=?

∫(0~π/2) √(1 - sin2x) dx
= ∫(0~π/2) √(sin²x - 2sinxcosx + cos²x) dx
= ∫(0~π/2) √(sinx - cosx)² dx
= ∫(0~π/2) |sinx - cosx| dx
解sinx - cosx = 0,x∈[0,π/2]
得tanx = 1 => x = π/4
当x < π/4,sinx - cosx < 0
当x > π/4,sinx - cosx > 0
= ∫(0~π/4) [- (sinx - cosx)] dx + ∫(π/4~π/2) (sinx - cosx) dx
= sinx + cosx |(0~π/4) + (- cosx - sinx) |(π/4~π/2)
= (√2 - 1) - (1 - √2)
= 2√2 - 2