含π+α诱导类型三角函数的不定积分

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：