例4求x2V1-x2 解 =sint,dx=cosidix=0.-0:x=1,= π d=sin't-sin'icostdt sin'tcos disin2dr ddcos4rd 4J0 π (t 8 -4sn42)= 2 4 16 11 11 例 4 求 1 2 2 0 x x dx 1−  解 令 sin , cos , 0, 0; 1, 2 x t dx tdt x t x t  = = = = = = 1 2 2 2 2 2 0 0 x x dx t t tdt 1 sin 1 sin cos  − = −   2 2 2 2 2 0 0 1 sin cos sin 2 4 t tdt tdt   = =   2 2 2 0 0 0 1 1 cos4 1[ cos4 ] 4 2 8 t dt dt tdt    − = = −    1 1 ( sin 4 ) 2 2 8 4 16 0 0 t t    = − =