含1.5π+α诱导类型三角函数的不定积分

时间:2024-10-12 11:25:16

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,∫ta荏鱿胫协n(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)=-cosαcos(3π/2+α)=sinαtan(3π/2+α)=-cotαcot(3π/2+α)=-tanαsec(3π/2+α)=-cscαcsc(3π/2+α)=secα

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)蟠校盯昂dα=∫[sin(3π/2+α)d(3π/2+α)/ cos(3π/2+α)]=-迟嗵莘啃∫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)蟠校盯昂dα=∫d(3π/2+α)/ cos(3π/2+α)=∫cos(3π/2+α)d(泌驾台佐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)蟠校盯昂dα=∫d(3π/2+α)/ sin(3π/2+α)=∫sin(3π/2+α)d(泌驾台佐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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