Question

求这三道题的详细过程

Answer 1

换元t=x-1，
=∫(-1到1)(t+1)/(t²+1)²d(t+1)
定积分偶倍奇零性质
=2∫(0到1)1/(t²+1)²dt
继续换元t=tanu
=2∫(0到π/4)1/(sec²u)²dtanu
=∫2cos²udu
=∫cos2u+1du
=sin2u/2+u
=1/2+π/4

追答

二倍角公式降次
=2∫(0到π/2)(cos2x+1)²dx
=∫cos4x+1+4cos2x+2dx
=sin4x/4+2sin2x+3x
=3π/2
学了牛莱公式更简单，直接=8*3/4*1/2*π/2=3π/2

也应用了定积分偶倍奇零性质，
=2∫(0到π/2)√cosx*sinxdx
=-2∫(cosx)^(1/2)dcosx
=(-4/3)(cosx)^(3/2)
=4/3