含2kπ+α诱导类型三角函数的不定积分

2025-10-23 06:33:52

1、sin(2kπ+α)=sin α

cos(2kπ+α)=cos α

tan(2kπ+α)=tan α

cot(2kπ+α)=cot α

sec(2kπ+α)=sec α

csc(2kπ+α)=csc α

含2kπ+α诱导类型三角函数的不定积分

1、∫sin(2kπ+α)dα

=∫sin(2kπ+α)d(2kπ+α)

=-cos(2kπ+α)+c

=-cosα+c

2、图例解析如下:

含2kπ+α诱导类型三角函数的不定积分

1、∫cos(2kπ+α)dα

=∫cos(2kπ+α)d(2kπ+α)

=sina(2kπ+α)+c

=sinα+c

2、图例解析如下:

含2kπ+α诱导类型三角函数的不定积分

1、 ∫tan(2kπ+α)dα

=∫[sin(2kπ+α) d(2kπ+α)/ cos(2kπ+α)]

=-∫d cos(2kπ+α)/cos(2kπ+α)

=-ln|cos(2kπ+α)|+c

=-ln|cosα|+c

 

2、图例解析如下:

含2kπ+α诱导类型三角函数的不定积分

1、∫ctg(2kπ+α)dα

=∫[cos(2kπ+α) d(2kπ+α)/ sin(2kπ+α)]

=∫d sin(2kπ+α)/sin(2kπ+α)

=ln|sin(2kπ+α)|+c

=ln|sinα|+c

2、图例解析如下:

含2kπ+α诱导类型三角函数的不定积分

1、∫sec(2kπ+α)dα

=∫dα/ cos(2kπ+α)

=∫d(2kπ+α)/ cos(2kπ+α)

=∫cos(2kπ+α)d(2kπ+α)/ [cos(2kπ+α)]^2

=∫dsin(2kπ+α)/ {1-[sin(2kπ+α)]^2}

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

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

=(1/2)ln{[1+sin(2kπ+α)]/ [1-sin(2kπ+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:

含2kπ+α诱导类型三角函数的不定积分

1、∫csc(2kπ+α)dα

=∫dα/ sin(2kπ+α)

=∫d(2kπ+α)/ sin(2kπ+α)

=∫sin(2kπ+α)d(2kπ+α)/ [sin(2kπ+α)]^2

=-∫dcos(2kπ+α)/ {1-[cos(2kπ+α)]^2}

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

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

=-(1/2)ln{[1+cos(2kπ+α)]/ [1-cos(2kπ+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+ctga|+c

2、图例解析如下:

含2kπ+α诱导类型三角函数的不定积分

声明:本网站引用、摘录或转载内容仅供网站访问者交流或参考,不代表本站立场,如存在版权或非法内容,请联系站长删除,联系邮箱:site.kefu@qq.com。
猜你喜欢