上式反过来看，∫fo(x)()=fuh为凑分法公 式。 例1求(x2+1xd 解r+=2cx+x+1y达 令x2+1=t,x=0,t=1,x=1,t=2 er+h-rh=g-点 或刘ra2+广ar--gr+r6 8
8 上 式反过来看， ( ( )) ( ) ( ) b a f x x dx f t dt = 为凑分法公 式。 例 1 求 1 2 3 0 ( 1) x xdx + 解 1 1 2 3 2 3 2 0 0 1 ( 1) ( 1) ( 1) 2 x xdx x x dx + = + + 令 2 x t x t x t + = = = = = 1 , 0, 1; 1, 2 1 2 2 3 3 4 0 1 1 1 15 2 ( 1) 2 8 8 1 x xdx t dt t + = = = 或 1 1 2 3 2 3 2 2 4 0 0 1 1 15 1 ( 1) ( 1) ( 1) ( 1) 2 8 8 0 x xdx x d x x + = + + = + =
例2求2 cos'xsin xdx cos'xsinxdcos(cos x) -2 1-6 0 9
9 例 2 求 2 5 0 cos sin x xdx 解 2 2 5 5 0 0 cos sin cos (cos ) x xdx x x dx = − 2 5 6 0 1 1 cos cos cos 2 6 6 0 xd x x = − = − =
例2求∫2 cos'xsinxdx [cos'xsin xdx=-[cosx(cosx)dx 2eo8 dcos6ow✉38 解了中n=矿i nx) =2v1+Inx 10
10 例 2 求 2 5 0 cos sin x xdx 解 2 2 5 5 0 0 cos sin cos (cos ) x xdx x x dx = − 2 5 6 0 1 1 cos cos cos 2 6 6 0 xd x x = − = − = 例 3 求 3 1 1 1 ln e dx x x + 解 3 1 1 1 ln e dx x x + 3 1 1 (ln 1) 1 ln e d x x = + + 3 2 1 ln 2 1 e = + = x
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 = − =