题目

∫(0,3 π)根号下1-cosx dx=

提问时间：2020-07-30

答案

=根号2·∫(0,3 π)根号下[(1-cosx)/2] dx
=根号2·∫(0,3 π)根号下[sin^2 (x/2)] dx
=根号2·∫(0,3 π) |sin (x/2)| dx
=2根号2·[∫(0,2 π) sin (x/2) d(x/2) - ∫(2π,3π) sin (x/2) d(x/2)]
=2√2·[ -cos(x/2) |(0,2 π) + cos(x/2) dx |(2π,3π)]
=2√2×(2 -1)
=2√2