含-α诱导类型三角函数的不定积分

2025-09-28 14:13:05

1、sin(-α)=-sin α

cos(-α)=cos α

tan(-α)=-tan α

cot(-α)=-cot α

sec(-α)=sec α

csc(-α)=-csc α

2、图例解析如下:

含-α诱导类型三角函数的不定积分

1、∫sin(-α)dα

=-∫sin(-α)d(-α)

=cos(-α)+c

=cosα+c

2、图例解析如下:

含-α诱导类型三角函数的不定积分

1、 

∫cos(-α)dα

=-∫cos(-α)d(-α)

=-sin(-α)+c

=sinα+c

2、图例解析如下:

含-α诱导类型三角函数的不定积分

1、 

∫tan(-α)dα

=-∫tan(-α)d(-α)

=-∫[sin(-α) d(-α)/ cos(-α)]

=∫d cos(-α)/cos(-α)

=ln|cos(-α)|+c

=ln|cosα|+c

2、图例解析如下:

含-α诱导类型三角函数的不定积分

1、∫cot(-α)dα

=-∫cot(-α)d(-α)

=-∫[cos(-α) d(-α)/ sin(-α)]

=-∫d sin(-α)/sin(-α)

=-ln|sin(-α)|+c

=-ln|sinα|+c

2、图例解析如下:

含-α诱导类型三角函数的不定积分

1、∫sec(-α)dα

=-∫sec(-α)dα

=-∫d(-α)/ cos(-α)

=-∫cos(-α)d(-α)/ [cos(-α)]^2

=-∫dsin(-α)/ [1-(sin(-α))^2}

=-∫dsin(-α)/ [(1-sin(-α))(1+ sin(-α))]

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

=-(1/2)ln{[1+sin(-α)]/ [1-sin(-α)]}+c

=-(1/2)ln[(1+sin(-α))/(1-sin(-α))]+c

=-(1/2)ln[(1+sin(-α))^2/(cos(-α))^2]+c

=-ln|(1+sin(-α))/cos(-α)|+c

=-ln|(1-sinα)/cosα|+c

=-ln|secα-tanα|+c

2、图例解析如下:

含-α诱导类型三角函数的不定积分

1、∫csc(-α)dα

=-∫csc(-α)d(-α)

=-∫d(-α)/ sin(-α)

=-∫sin(-α)d(-α)/ [sin(-α)]^2

=∫dcos(-α)/ [1-(cos(-α))^2]

=∫dcos(-α)/ [(1-cos(-α))(1+ cos(-α))]

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

=(1/2)ln[(1+cos(-α))/ (1-cos(-α))]+c

=(1/2)ln[(1+cos(-α))^2/(sin(-α))^2]+c

=ln|(1+cos(-α))/sin(-α)|+c

=ln|(1+cosα)/sinα|+c

=ln|cscα+cota|+c

2、图例解析如下:

含-α诱导类型三角函数的不定积分

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