例2计算「 01+c0s2x 解1+c0s2x=2c0s2x, r xix tan 01+cos 2x Jo 2 cosx Kx tanx]o-aJo tan xdx 2 π—8 In sec x πIn2 841 cos2 2cos , 2  + x = x   +  4 0 1 cos 2x xdx   = 4 0 2 2cos x xdx d( x) x tan 2 4 0  =   4 0 tan 2 1  = x x tan xdx 2 1 4 0  −   4 0 lnsec 2 1 8  −  = x . 4 ln2 8 −  = 例2 计算 . 1 cos 2 4 0  + x xdx 解