含1.5π-α诱导类型三角函数的不定积分

1、∫sin(3π/2-α)dα

=-∫sin(3π/2-α)d(3π/2-α)

=cos(3π/2-α)+c

=-cos(π/2-α)+c

=-sinα+c

2、图例解析如下：

1、∫sin(3π/2-α)dα

=-∫sin(3π/2-α)d(3π/2-α)

=cos(3π/2-α)+c

=-cos(π/2-α)+c

=-sinα+c

2、图例解析如下：

1、∫cos(3π/2-α)dα

=-∫cos(3π/2-α)d(3π/2-α)

=-sin(3π/2-α)+c

=sin(π/2-α)+c

=cosα+c

2、图例解析如下：

1、∫tan(3π/2-α)dα

=-∫[sin(3π/2-α) d(3π/2-α)/ cos(3π/2-α)]

=∫d cos(3π/2-α)/cos(3π/2-α)

=ln|cos(3π/2-α)|+c

=ln|sinα|+c

2、图例解析如下：

1、∫cot(3π/2-α)dα

=-∫[cos(3π/2-α) d(3π/2-α)/ sin(3π/2-α)]

=-∫d sin(3π/2-α)/sin(3π/2-α)

=-ln|sin(3π/2-α)|+c

=-ln|cosα|+c

2、图例解析如下：

1、∫sec(3π/2-α)dα

=-∫d(3π/2-α)/ cos(3π/2-α)

=-∫cos(3π/2-α)d(3π/2-α)/ [cos(3π/2-α)]^2

=-∫dsin(3π/2-α)/ {1-[sin(3π/2-α)]^2}

=-∫dsin(3π/2-α)/ {[1-sin(3π/2-α)][1+ sin(3π/2-α)]}

=-(1/2){∫dsin(3π/2-α)/ [1-sin(3π/2-α)]+∫dsin(3π/2-α)/ [1+sin(3π/2-α)]}

=-(1/2)ln{[1+sin(3π/2-α)]/ [1-sin(3π/2-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：

1、∫csc(3π/2-α)dα

=-∫d(3π/2-α)/ sin(3π/2-α)

=-∫sin(3π/2-α)d(3π/2-α)/ [sin(3π/2-α)]^2

=∫dcos(3π/2-α)/ {1-[cos(3π/2-α)]^2}

=∫dcos(3π/2-α)/ {[1-cos(3π/2-α)][1+ cos(3π/2-α)]}

=(1/2){∫dcos(3π/2-α)/ [1-cos(3π/2-α)]+∫dcos(3π/2-α)/ [1+cos(3π/2-α)]}

=(1/2)ln{[1+cos(3π/2-α)]/ [1-cos(3π/2-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+tana|+c

2、图例解析如下：