1、sin(3π/2+α)=-cosα
cos(3π/2+α)=sinα
tan(3π/2+α)=-cotα
cot(3π/2+α)=-tanα
sec(3π/2+α)=-cscα
csc(3π/2+α)=secα
2、图例解析如下：
1、∫sin(3π/2+α)dα
=∫sin(3π/2+α)d(3π/2+α)
=-cos(3π/2+α)+c
=cos(π/2+α)+c
=-sinα+c
2、图例解析如下：
1、∫cos(3π/2+α)dα
=∫cos(3π/2+α)d(3π/2+α)
=sin(3π/2+α)+c
=-sin(π/2+α)+c
=-cosα+c
2、图例解析如下：
1、∫tan(3π/2+α)dα
=∫[sin(3π/2+α) d(3π/2+α)/ cos(3π/2+α)]
=-∫d cos(3π/2+α)/cos(3π/2+α)
=-ln|cos(3π/2+α)|+c
=-ln|sinα|+c
2、图例解析如下：
1、
∫cot(3π/2+α)dα
=∫[cos(3π/2+α) d(3π/2+α)/ sin(3π/2+α)]
=∫d sin(3π/2+α)/sin(3π/2+α)
=ln|sin(3π/2+α)|+c
=ln|cosα|+c
2、图例解析如下：
1、∫sec(3π/2+α)dα
=∫d(3π/2+α)/ cos(3π/2+α)
=∫cos(3π/2+α)d(3π/2+α)/ [cos(3π/2+α)]^2
=∫dsin(3π/2+α)/ {1-[sin(3π/2+α)]^2}
=∫dsin(3π/2+α)/ {[1-sin(3π/2+α)][1+ sin(3π/2+α)]}
=(1/2){∫dsin(3π/2+α)/ [1-sin(3π/2+α)]+∫dsin(3π/2+α)/ [1+sin(3π/2+α)]}
=(1/2)ln{[1+sin(3π/2+α)]/ [1-sin(3π/2+α)]}+c
=(1/2)ln[(1+cosα)/(1-cosα)]+c
=(1/2)ln[(1+cosα)^2/(sinα)^2]+c
=ln|(1+cosα)/sinα|+c
=ln|cscα+cotα|+c
2、图例解析如下：
1、∫csc(3π/2+α)dα
=∫d(3π/2+α)/ sin(3π/2+α)
=∫sin(3π/2+α)d(3π/2+α)/ [sin(3π/2+α)]^2
=-∫dcos(3π/2+α)/ {1-[cos(3π/2+α)]^2}
=-∫dcos(3π/2+α)/ {[1-cos(3π/2+α)][1+ cos(3π/2+α)]}
=-(1/2){∫dcos(3π/2+α)/ [1-cos(3π/2+α)]+∫dcos(3π/2+α)/ [1+cos(3π/2+α)]}
=-(1/2)ln{[1+cos(3π/2+α)]/ [1-cos(3π/2+α)]}+c
=-(1/2)ln[(1-sinα)/(1+sinα)]+c
=-(1/2)ln[(1-sinα)^2/(cosα)^2]+c
=-ln|(1-sinα)/cosα|+c
=-ln|secα-tana|+c
2、图例解析如下：
