不知道你的题目是个啥呢,能具体说说不,我来解解看就是sin(x/2+派/4)=cos(派/4 -x/2) 那么就可以得到原式=sin(派/4 -x/2) /cos(派/4 -x/2) *sinx =tan(派/4 -x/2) *sinx
...... = (π/2)(1/2)∫<0, 1>ln(1+p^2)d(1+p^2)
= (π/4){[(1+p^2)ln(1+p^2)]<0, 1> - ∫<0, 1>(1+p^2) 2pdp/(1+p^2)}
= (π/4){[(1+p^2)ln(1+p^2)]<0, 1> - ∫<0, 1>2pdp}
= ......
分步积分啊……