设f(x)=∫sint/π-tdt(0→x),求∫f(x)dx(0→π)

Question

设f(x)=∫sint/π-tdt(0→x),求∫f(x)dx(0→π)

Answer 1

记f'(x) = sinx/(π - x)
∫(0~π) f(x) dx
= xf(x) - ∫(0~π) xf(x)' dx、
= πf(π) - ∫(0~π) x · sinx/(π - x) dx
= π∫(0~π) sint/(π - t) dt - ∫(0~π) xsinx/(π - x) dx
= π∫(0~π) sinx/(π - x) dx - ∫(0~π) xsinx/(π - x) dx
= ∫(0~π) (πsinx - xsinx)/(π - x) dx
= ∫(0~π) (π - x)sinx/(π - x) dx
= ∫(0~π) sinx dx
= - cosx |_(0~π)
= - (- 1 - 1)
= 2