1/cos2x的不定积分
问题描述：
1/cos2x的不定积分
答
∫ dx/cos2x= ∫ sec2x dx= ∫ sec2x * (sec2x+tan2x)/(sec2x+tan2x) dx= (1/2)∫ (sec²2x+sec2xtan2x)/(sec2x+tan2x) d(2x)= (1/2)∫ d(tan2x+sec2x)/(sec2x+tan2x)= (1/2)ln|sec2x+tan2x| + C