大一新生求助各位积佬

∫ sinx.cosx/(sinx+cosx) dx
=(1/2)∫ sin2x/(sinx+cosx) dx
=-(1/4)∫ dcos2x/(sinx+cosx)
=-(1/4)[cos2x/(sinx+cosx)]- (1/4)∫ cos2x.(cosx-sinx)/(sinx+cosx)^2 dx
*****Note: cos2x = (cosx-sinx)(cosx+sinx)
=-(1/4)[cos2x/(sinx+cosx)]- (1/4)∫ (cosx-sinx)^2/(sinx+cosx) dx
=-(1/4)[cos2x/(sinx+cosx)]- (1/4)∫ (1-2sinx.cosx)/(sinx+cosx) dx
=-(1/4)[cos2x/(sinx+cosx)]- (1/4)∫ dx/(sinx+cosx) +(1/2)∫ sinx.cosx/(sinx+cosx) dx
=-(1/8)[cos2x/(sinx+cosx)]- (1/8)∫ dx/(sinx+cosx)
=-(1/8)[cos2x/(sinx+cosx)]- [1/(8√2)]∫ dx/sin(x+π/4)
=-(1/8)[cos2x/(sinx+cosx)]- [1/(8√2)]∫ csc(x+π/4) dx
=-(1/8)[cos2x/(sinx+cosx)]- [1/(8√2)]ln|csc(x+π/4)-cot(x+π/4)| + C