第2个回答 推荐于2017-11-18
(1)
∫(0->4) dx/√(2x+1)
=(1/2)∫(0->4) d(2x+1)/√(2x+1)
= [√(2x+1)]|(0->4)
=3-1
=2
(2)
∫(0->1) x(x^2+1)^16 dx
=(1/2) ∫(0->1) (x^2+1)^16 d(x^2+1)
=(1/34)[ x^2+1)^17] |(0->1)
=(2^17 -1 ) /34
(3)
∫(0->π/2) √[cosx - (cosx)^3 ] dx
=∫(0->π/2) sinx.√cosx dx
=-∫(0->π/2) √cosx dcosx
=- (2/3) [ (cosx)^(3/2) ]|(0->π/2)
=2/3本回答被网友采纳