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