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