tan't -sec2 tdt -∫eos,iast-jd-7uosi cost+ 1 cost 3.含√x2-a2型(a>0) V1+x2 方法：令x=secl,i∈(0，) 8
8 3 3 2 2 3 2 3 tan sec (1 ) (1 tan ) x t dx tdt x t = + + 3 2 sin cos t dt t = 2 2 sin cos cos t d t t = − 2 2 cos 1 cos cos t d t t − = 2 1 (1 ) cos cos d t t = − 2 2 1 1 cos 1 cos 1 t c x c t x = + + = + + + + 3．含 2 2 x a − 型( 0) a 方法：令 sec , (0, ) 2 x a t t = t x 2 1+ x 1
例1.求 1 x-d dx(a>0) 解令x=asect,,dk=asecttantdt 1w2-a asecttant dt =sectdt In sect tant+c 2-a2 =In x.vx2-a2 +C1 a 2 -Inx+vx"-a"te 9
9 例 1．求 2 2 1 dx a( 0) x a − 解 令x a t dx a t tdt = = sec , sec tan 2 2 2 2 2 1 sec tan sec a t t dx dt x a a t a = − − = sectdt 1 2 2 1 2 2 ln sec tan ln ln t t c x x a c a a x x a c = + + − = + + = + − + t x 2 2 x a − a
例2 _dx 解令x=sect,d=secttantdt 1-- 1 =t+c=arccos-+c X 4.含ar+b型 方法：令ax+b=t 例1 解令√X=t,x=t,dx=2tdh 10
10 例 2．求 2 1 1 dx x x − 解 令x t dx t tdt = = sec , sec tan 2 2 1 sec tan 1 sec sec 1 t t dx dt dt x x t t = = − − 1 t c c arccos x = + = + 4．含k ax b + 型 方法：令k ax b t + = 例 1.求 1 1 dx + x 解 令 2 x t x t dx tdt = = = , , 2
1+s-J2a咖=可a =2-1+7h=2-l+0+c =2[V-ln(1+Vx)]+c 例2树0 解令=t,x=t,d=6tdi ia-aa-ah-,a -6l-17h=6-am+e =6(/x-arctanx)+c 11
11 1 1 1 1 2 2 1 1 1 t dx tdt dt x t t + − = = + + + 1 2 (1 ) 2( ln 1 ) 1 dt t t c t = − = − + + + = − + + 2[ ln(1 )] x x c 例 2.求 3 1 (1 ) dx + x x 解 令6 6 5 x t x t dx t dt = = = , , 6 5 2 3 2 3 2 1 6 6 (1 ) (1 ) 1 t t dx dt dt x x t t t = = + + + 2 1 6 (1 ) 6( arctan ) 1 dt t t c t = − = − + + 6 6 = − + 6( arctan ) x x c