设∫f(tx)dt=f(x)+sinx,求连续函数f(x),积分上下限是0到1

题目修正：∫[0,1] f(tx) dt = f(x) + xsinx
令u = tx,du = xdt => dt = du/x
当t = 0,u = 0；当t = 1,u = x
∫[0,1] f(tx) dt = (1/x)∫[0,x] f(u) du = f(x) + xsinx
∫[0,x] f(u) du = xf(x) + x²sinx,两边求导
d/dx ∫[0,x] f(u) du = d/dx xf(x) + d/dx x²sinx
f(x) = xf'(x) + f(x) + x²cosx + 2xsinx
xf'(x) = -x²cosx - 2xsinx
f'(x) = -xcosx - 2sinx
f(x) = -∫ xcosx dx - 2∫ sinx dx
= -∫ x d(sinx) - 2∫ sinx dx
= -xsinx + ∫ sinx dx - 2∫ sinx dx
= -xsinx - ∫ sinx dx
= -xsinx + cosx + C