请问：d/dx(∫sin^2xdx）,即在0-x之间sin^2tdt的定积分的微分怎么算?(定积分的符号打不上）

Question

请问：d/dx(∫sin^2xdx）,即在0-x之间sin^2tdt的定积分的微分怎么算?(定积分的符号打不上）

Answer 1

那就先求积分,后求导数吧
d/dx ∫(sin²t)dt
=d/dx (1/2)∫(1-cos2t)dt
=d/dx (1/2)[∫dt-(1/2)∫cos2td(2t)]
=d/dx (1/2)[t-(1/2)*sin2t+C]
=d/dx (1/2)[x-(1/2)sin2x+C]
=(1/2) d/dx[x-(1/2)sin2x]
=(1/2)[1-(1/2)*cos2x*2]
=(1/2)(1-cos2x)