Question

∫0-3л 根号下（1-cosx）dx怎么求〒_〒

Answer 1

=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