∫cosx/(sinx+cosx)dx请用换元法解答在【0 π/2】上的定积分.

cosx = A(sinx + cosx) + B(cosx - sinx)cosx = (A-B)sinx + (A+B)cosxA=B,A+B=12A=1=>A=1/2,B=1/2∫ cosx/(sinx+cosx) dx= (1/2)∫ dx + (1/2)∫ (cosx-sinx)/(sinx+cosx) dx,令u=sinx+cosx,du=(cosx-sinx)dx= x/2 ...