含2π+α诱导类型三角函数的不定积分

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：