三角函数基本知识
不定积分基本知识
∫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
∫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
图例解析如下：
∫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
图例解析如下：
∫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
图例解析如下：
∫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
图例解析如下：
∫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
图例解析如下：
∫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
图例解析如下：