求下列不定积分：
(1)∫[cos2x/(sinx-cosx)]dx
(2)∫[(1+sin2x)/sinx+cosx)]dx．
【正确答案】：(1)∫[cos2x/(sinx-cosx)]dx=∫[(cos2x-sin2x)/(sinx-cosx)]dx =(-sinx-cosx)dx =cosx-sinx+C． (2)∫[(1+sin2x)/(sinx+cosx)]dx=∫[(1+2sinxcosx)/(sinx+cosx)]dx =∫[(sinx+cosx)2/(sin+cosx)]dx =∫(sinx+cosx)dx =∫(sinx+cosx)dx =-cosx+sinx+C．
求下列不定积分：(1)∫[cos2x/(sinx-cosx)]dx(2)∫[(1+sin2x)/sinx+cosx)]dx．
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