32 cost+—cos +(4-x2)-4-x2)2 dx(a>0 解.令x=asec jrra-dra>o)=atan a sec tant=a tan d =a tani-at- (1+e e(1+e)d= 1rd(1 = arcsin e-√1- d(a>0) v2a=x d全a=x2「n=hn全m=y2nsmt8 Ba tdt 2a 8a2〔-c9202 dt= 2a(1-2 cos 2t+cos 2t)dt =2at-2a- sin 2t+ 2a 1+cos 4t dt= 3a t-2a sin 2t +-sin 4t+c =3a2t-4a sin t cost+a2 sin t cost (1-2sin2t)+c =3att-3a sin tcost-2a' sint cost+c 2a2 Vav 2 2a v 2av 2 3a arcsin 3a+x a)√x(2a-x)+c 十.求下列不定积分 2-sin x 2 +coSx 2-sin x 1dr+rd(2 cos cosx 2+cos x 2+cosx 2dt 令m2=:2-1+2-+ln12 2dt cOSx 2「 +In 2+cosx= - t + t + c = - x - - x + c 2 3 2 2 5 3 5 2 (4 ) 3 4 (4 ) 5 1 cos 5 32 cos 3 32 2. Ú > - ( 0 ) 2 2 dx a x x a 解. 令 x = a sec t Ú Ú Ú > = = = - + - a t t a tdt a t at c a t a t dx a x x a sec tan tan tan sec tan ( 0 ) 2 2 2 = c x a x - a - a arccos + 2 2 3. dx e e e x x x Ú - + 2 1 (1 ) 解. = - + Ú d e e e x x x 2 1 (1 ) Ú - dx e e x x 2 1 + dx e e x x Ú - 2 2 1 = Ú - x x e de 2 1 - dx e d e x x Ú - - 2 2 1 (1 ) 2 1 = e e c x x - - + 2 arcsin 1 4. Ú - dx a x x x 2 (a > 0) 解. Ú - dx a x x x 2 令u = x Ú - du a u u 2 4 2 2 令u = 2a sin t Ú a tdt 2 4 8 sin = Ú Ú = - + - dt a t t dt t a 2 (1 2 cos2 cos 2 ) 4 (1 cos 2 ) 8 2 2 2 2 = t c a dt a t a t t a t a t a = - + + + - + Ú sin 4 4 3 2 sin 2 2 1 cos 4 2 2 sin 2 2 4 2 2 2 2 2 =3 a t - 4 a sin t cost + a sin t cost (1 - 2 sin t ) + c 2 2 2 2 =3 a t - 3 a sin t cost - 2 a sin t cost + c 2 2 2 3 = c a a x a x a x a a a x a x a a x a + - - - - 2 2 2 2 2 2 2 2 3 2 3 arcsin 2 2 2 = x a x c a x a x a - + + - (2 ) 2 3 2 3 arcsin 2 十. 求下列不定积分: 1. Ú + - dx x x 2 cos 2 sin 解. Ú Ú Ú + + + + = + - x d x dx x dx x x 2 cos (2 cos ) 2 cos 1 2 2 cos 2 sin t x = 2 令tan Ú Ú + + + + + = + - + + ln | 2 cos | 3 2 ln | 2 cos | 2 1 1 2 1 2 2 2 2 2 2 x t dt x t t t dt