设f'(x)=cosx/(1+sinx^2),且f(0)=0,则∫f'(x)/(1+f(x)^2)dx=
设f'(x)=cosx/(1+sinx^2),且f(0)=0,则∫f'(x)/(1+f(x)^2)dx=
设f(x)导数在【-1,1】上连续,且f(0)=1,计算∫【f(cosx)cosx-f‘(cosx)sin^2x】dx(
100分求高数积分题设f(x)在[-π,π]上连续 且f(x)=x/(1+(cosx)^2)+∫ f(x)sinX dx
设f(x)在【0,1】上连续.证明∫(π/2~0)f(cosx)dx=∫(π/2~0)f(sinx)dx
已知f(cosx)=(sinx)∧2,则∫f(x-1)dx=?
设函数f(x)=sinx-cosx+x+1,0
若∫f(x)dx=1/2x^2+C 则∫f(sinx)dx= -cosx+c
,设f(x)是连续函数,且f(x)=x+2∫[1,0]f(t)dt ,则∫[1,0]f(x)dx=?
设f(x)为连续函数,且满足f(x)=3x^2-x∫(1,0)f(x)dx求f(x)
f(x) = x - ∫(0~π) f(x) * cosx dx f'(x) = 1
设2f(x)cos x=d/dx [f(x)]²,f(0)=1,则f(x)=
设f(x)∈C[0,1],证明∫(π,0)*x*f(sinx)dx =π/2*∫(π,0)*f(sinx)dx