含1.5π+α诱导类型三角函数的不定积分

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：

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：

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：

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：