Question

求不定积分：∫ cosx/(sinx+cosx) dx

Answer 1

∫ cosx/(sinx+cosx) dx
=∫ cosx( cosx- sinx)/((cosx)^2-(sinx)^2 ) dx
= (1/2)∫ [(cos2x-1) - sin2x]/ (cos2x) dx
= (1/2)∫ ( 1- tan2x - sec2x) dx
=(1/2) ( x + (1/2)ln|cos2x| - (1/2)log|sec2x+tan2x| ) + C

Answer 2

∫cosx/(sinx+cosx) dx
= (1/2)∫[(cosx+sinx)+(cosx-sinx)]/(sinx+cos)] dx
= (1/2)∫ dx + (1/2)∫(cosx-sinx)/(sinx+cosx) dx
= x/2 + (1/2)∫d(sinx+cosx)/(sinx+cosx)
= (1/2)(x+ln|sinx+cosx|) + C

参考:
A=∫cosx/（sinx+cosx)dx
B=∫sinx/(sinx+cosx)dx
A+B=∫(cosx+sinx)/(sinx+cosx)dx =∫dx =x+c (1)
A-B=∫(cosx-sinx)/(sinx+cosx)dx =∫(d(cosx+sinx)/(sinx+cosx)=ln(cosx+sinx)+c (2)
[(1)+(2)]/2得：
A=∫cosx/(sinx+cosx)dx =x/2+1/2*ln(cosx+sinx)+c