例5求[ n sin 2xdx 解令2x=,「 In sin2xt=1(2 In sin tdi 2 Insin 2xdx=4 In(2 sin x cos x)dx J(n2+ In sin x+ In cos x)dx In 2+4Insin xdx+2In sin xdx 4 T In 2+2Insin xdx ==In 2+21 In 2lnsin 2 . 4 0  求 xdx 解 令 2x = t, lnsin . 2 1 lnsin 2 2 0 4 0    xdx = tdt   = 4 0 I lnsin 2xdx   = 4 0 ln(2sin x cos x)dx   = + + 4 0 (ln 2 lnsin x lncos x)dx      + +  = 2 4 4 0 ln 2 lnsin lnsin 4 xdx xdx   +  = 2 0 ln2 lnsin 4 xdx ln 2 2I 4 +  = ln2. 4   I = − 例 5