王 例4计算∫ dx.(>0) x+√a-x 庄解令x=asmn,a=acb, T V=a→t= x=0→t=0, 2 T 原式=[2 a cos t dt Jo asin t+√a2(1-sin2) cos t cost→sint dt 21+。dr Jo sint+cost 2 Jo sint+ cos t T +-lnsint+ cos t 222 4 上页例4 计算 解   + − a dx a x a x 0 2 2 . ( 0) 1 令 x = asint, x = a , 2   t = x = 0  t = 0, dx = acostdt, 原式   + − = 2 0 2 2 sin (1 sin ) cos dt a t a t a t   + = 2 0 sin cos cos dt t t t         + − = + 2 0 sin cos cos sin 1 2 1 dt t t t t   2 0 lnsin cos 2 1 2 2 1  + +  =  t t . 4  =