∫(0,π){(xsinx)/[(1+(cosx)^2]}dx=(π^2)/4 ∫(0,π/2)[(sinx)^2/(sinx+cosx)]dx=(1/√2)ln(1+√2)

算出来了，太难打了第一题∫(0,π){(xsinx)/[(1+(cosx)^2]}dx=-∫(0,π)xd(arctan(cosx)dx分步积分=-xd(arctan(cosx)dx+∫(0,π)arctan(cosx)dx=(π^2)/4 +∫(0,π)arctan(cosx)dx令cosx=t换元后∫(0,π)arctan(cosx)dx=-∫(-1,1)(arctant)/(1-t^2)^(1/2)dt=02)利用sinx^2=(1/2)(1-cos2x)代入后拆成两项分别积分即证。过程不写了最后一步积分为(√2/4)ln│csc(x+π/4)-cot(x+π/4)│，x:(0,π/2)=(1/√2)ln(1+√2)