高数已知f（x）的原函数（1＋sinx）lnx,求定积分：上限派∏,下限派∏/2,求xdf（x）

Question

高数已知f（x）的原函数（1＋sinx）lnx,求定积分：上限派∏,下限派∏/2,求xdf（x）

Answer 1

f(x) = [(1 + sinx)lnx]' = (sinx + xcosxlnx + 1)/x
∫(π/2~π) x df(x)
= xf(x) |_π/2^π - ∫(π/2~π) f(x) dx
= (sinx + xcosxlnx + 1) |_π/2^π - (1 + sinx)lnx |_π/2^π
= [(0 - πlnπ + 1) - (1 + 0 + 1)] - [(1 + 0)lnπ - (1 + 1)ln(π/2)]
= (- πlnπ - 1) - (lnπ - 2lnπ + 2ln2)
= (1 - π)ln(π) - 1 - 2ln(2)