4)|edx=e+c 5)a2d a+c 5°(a2)y=lna:a2 na a>0.a≠1 a>0.a≠1 6 cos xdx =sin x+C 6°(sinx)=cosx 7) sin xdx=-cosx+C 7(cos x)=-sin x
5) 6) 7) 5 6 7 0, 1 ln1 d = + a a a C a a x x x x x = x + C cos d sin x x = - x + C sin d cos 0, 1 ( ) ln = a a a ' a a x x (sin x )' = cos x (cos x )' = -sin x 4 x x e x e C ( e ) ' = e x x = + 4) d
8)sec xdx=tgx+C 8(tgx)=sec x 9)」 csc x dx=-ctgx+C9°( (ctg))=-cc2x x -arcsinx+C 10(arcsin x)-1 √1-x arctgx+C11°( arctan) 1+x 1
10) 11) 10 11 9) 9 x x = - x + C csc d ctg 2 x ' x 2 (ctg ) = -csc x C xx = + - arcsin 1d 2 x C xx = + + arctg 1 d 2 2 11 (arcsin ) x x ' - = 2 1 1 (arctg ) x x ' + = 8) 8 x x = x + C sec d tg 2 x ' x 2 (tg ) = sec
积分公式的简单应用 例1.求J(x2+x2x 解:「(x2+x )dx x dx+xdx-2x'dx x3+x2++C
4. 积分公式的简单应用 例1. 求 + - x x x x x )d 2 ( 3 2 2 解: + - x x x x x )d 2 ( 3 2 2 - = x dx + x dx - 2 x dx 2 3 5 2 C x = x + x + +2 2 7 3 1 7 2 3 1
1+x+x 例2.求 x(1+x 2 1+x+x 解 x(1+x2) 1+x =arctan+In x +C
例2. 求 + + + x x x x x d (1 ) 1 2 2 解: + + = x x x x d 1 d 2 = arctan x + ln | x | +C + + + x x x x x d (1 ) 1 2 2
例3.求tan2xdx 解:|tan2xdx I(sec)dx=sec xdx- dx =tanx-x+c
例3. 求 解: tan xdx 2 (sec x 1)dx 2 = - = tan x - x +C tan xdx 2 = sec xdx - dx 2