含0.5π+α诱导类型三角函数的不定积分

 时间:2024-10-27 17:45:39

本经验介绍含0.5π+α诱珑廛躬儆导类型三角函数的不定积分,即求∫sin(0.5π+α)dα,∫cos(0.5π+α)dα,∫tan(0.5π+α)dα,∫cot烫喇霰嘴(0.5π+α)dα,∫sec(0.5π+α)dα,∫csc(0.5π+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0.5π+α三角函数的诱导公式

1、sin(π/2+α)=cosαcos(π/2+α)=−sinαtan(π/2+α)=-cotαcot(π/2+α)=-tanαsec(π/2+α)=-cscαcsc(π/2+α)=secα

2、图例解析如下:

含0.5π+α诱导类型三角函数的不定积分

2.sin(0.5π+α)

1、∫sin(π/2+α)dα=∫sin(π/2+α)d(π/2+α)=-cos(π/2+α)+c=sinα+c

2、图例解析如下:

含0.5π+α诱导类型三角函数的不定积分

3.cos(0.5π+α)

1、∫cos(π/2+α)dα=∫cos(π/2+α)d(π/2+α)=sin(π/2+α)+c=-cosα+c

2、图例解析如下:

含0.5π+α诱导类型三角函数的不定积分

4.tan(0.5π+α)

1、∫tan(π/2+α)d拿骛蟊痊α=∫[sin(π/2+α)d(π/2+α)/ cos(π/2+α)]=-∫d cos(π/2+α)/cos烫喇霰嘴(π/2+α)=-ln|cos(π/2+α)|+c=-ln|sinα|+c

2、图例解析如下:

含0.5π+α诱导类型三角函数的不定积分

5.cot(0.5π+α)

1、∫cot(π/2+α)dα=∫[cos(π/2+α)d(π/2+α)/ sin(π/2+α)]=∫d sin(π/2+α)/sin(π/2+α)=ln|sin(π/2+α)|+c=ln|cosα|+c

2、图例解析如下:

含0.5π+α诱导类型三角函数的不定积分

6.sec(0.5π+α)

1、∫sec(π/2+α)d拿骛蟊痊α=∫d(π/2+α)/ cos(π/2+α)=∫cos(π/2+α)d(π/2+α)/ [cos(π/2+珍提疮翘α)]^2=∫dsin(π/2+α)/ {1-[sin(π/2+α)]^2}=∫dsin(π/2+α)/ {[1-sin(π/2+α)][1+ sin(π/2+α)]}=(1/2){∫dsin(π/2+α)/ [1-sin(π/2+α)]+∫dsin(π/2+α)/ [1+sin(π/2+α)]}=(1/2)ln{[1+sin(π/2+α)]/ [1-sin(π/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0.5π+α诱导类型三角函数的不定积分

7.csc(0.5π+α)

1、∫csc(π/2+α)d拿骛蟊痊α=∫d(π/2+α)/ sin(π/2+α)=∫sin(π/2+α)d(π/2+α)/ [sin(π/2+珍提疮翘α)]^2=-∫dcos(π/2+α)/ {1-[cos(π/2+α)]^2}=-∫dcos(π/2+α)/ {[1-cos(π/2+α)][1+ cos(π/2+α)]}=-(1/2){∫dcos(π/2+α)/ [1-cos(π/2+α)]+∫dcos(π/2+α)/ [1+cos(π/2+α)]}=-(1/2)ln{[1+cos(π/2+α)]/ [1-cos(π/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0.5π+α诱导类型三角函数的不定积分
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