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)

=-sin(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：