y = ∫ [sinx, x]1/√(1+t+t²)dt, f(t) =1/√(1+t+t²)
直接求原函数,利用N-L公式,太复杂,需利用变上限积分求导:
dy/dx = f(x) - f(sinx) (sinx)' = f(x) - cosx f(sinx)
= 1/√(1+x+x²) - cosx / √(1+sinx+sin²x)
y=∫(从sin x到x)1/√(1+t+t∧2)dt,求dy/dx.大仙帮帮忙……
y=∫(从sin x到x)1/√(1+t+t∧2)dt,求dy/dx.大仙帮帮忙……
y=∫(从sin x到x)1/√(1+t+t∧2)dt,求dy/dx.
大仙帮帮忙……
y=∫(从sin x到x)1/√(1+t+t∧2)dt,求dy/dx.
大仙帮帮忙……
数学人气:475 ℃时间:2019-11-01 20:32:20
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