f(x) = x - ∫(0~π) f(x) * cosx dx f'(x) = 1

f(x) = x - ∫(0~π) f(x) * cosx dx f'(x) = 1 ∫f(x)dx=F(x)+C 求 ∫cosx f(sinx) dx 求定积分：∫f(x-1)dx,上限2,下限0,其中f(x)=cosx,若x>=0,f(x)=x+1,若x 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（ 若f(x)=e^x+2∫(0 1)f(x)dx 求f(x) 设f(x)-(cosx)^2=∫（下0上π/4）f(2x)dx,求∫（下0上π/2)f(x)dx. f(x)=e^x/x,求∫f'(x)dx/1+f^2(x)? 若∫f(x)dx=e^(-x)cosx+C,则f(x)= 设f'(x)=cosx/(1+sinx^2),且f(0)=0,则∫f'(x)/(1+f(x)^2)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 证明∫(-a,a)f(x)dx=∫(0,a)[f(x)+f(-x)]dx 求积分：∫(2,0)f(x)dx,其中f(x)=x,x=1