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