x3
(20)∫dx; 29+x
1x21912222 解 ∫dx=d(x)=(1 )d(x)=x 9ln(9+x)]+C. ∫∫2222229+x9+x9+x
1 (21)∫2dx; 2x 1x3
解 ∫2x2 1
1
2
1
221dx=∫1(2x 1)(2x+1)dx=111( )dx 2∫2x 12x+1 =∫212x 1d(2x 1) 12212∫2(2x+1) 2x+1122ln|2x 1
2x+1|+C. 1 =
(22)∫
解 ln|x 1| ln|x+1|+C=1dx; (x+1)(x 2)111111x 2dx=( )dx=(ln|x 2| ln|x+1|+C=ln|+C. ∫(x+1)(x 2)3∫x 2x+133x+1
(23)∫cos3xdx;
1 解 ∫cos3xdx=∫cos2xdsinx=∫(1 sin2x)dsinx=sinx sin3x+C. 3
(24)∫cos2(ωt+ )dt;
解 ∫cos2(ωt+ )dt=
(25)∫sin2xcos3xdx;
解 ∫sin2xcos3xdx=
x (26)∫cosxcosdx; 2
131131x 解 ∫cosxcosdx=∫(cosx+cosx)dx=sinx+sinx+C. 2222322111(sin5x sinx)dx= cos5x+cosx+C. ∫2102111[1+cos2(ωt+ )]dt=t+sin2(ωt+ )+C. 2∫24ω
(27)∫sin5xsin7xdx;
解 ∫sin5xsin7xdx= 111(cos12x cos2x)dx= sin12x+sin2x+C. 2∫244