含1.5π-α诱导类型三角函数的不定积分

时间:2024-10-16 01:13:33

本经验介绍含1.5π-α诱导类型三角函数的不定积分,即求∫sin(1.5π-α)dα,∫cos(1.5π-α)dα,∫tan(1.5π-α)dα,∫cot(1.5π-α)dα,∫sec(1.5π-α)dα,∫csc(1.5π-α)dα的步骤。

工具/原料

三角函数基本知识

不定积分基本知识

1.含1.5π-α三角函数的诱导公式

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:

含1.5π-α诱导类型三角函数的不定积分

2.sin(1.5π-α)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:

含1.5π-α诱导类型三角函数的不定积分

3.cos(1.5π-α)的不定积分

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:

含1.5π-α诱导类型三角函数的不定积分

4.tan(1.5π-α)的不定积分

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:

含1.5π-α诱导类型三角函数的不定积分

5.cot(1.5π-α)的不定积分

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:

含1.5π-α诱导类型三角函数的不定积分

6.sec(1.5π-α)的不定积分

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:

含1.5π-α诱导类型三角函数的不定积分

7.csc(1.5π-α)的不定积分

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:

含1.5π-α诱导类型三角函数的不定积分
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