三角函数基本知识
不定积分基本知识
sin(2kπ+α)=sin αcos(2kπ+α)=cos αtan(2kπ+α)=tan αcot(2kπ+α)=cot αsec(2kπ+α)=sec αcsc(2kπ+α)=csc α
∫sin(2kπ+α)dα=∫sin(2kπ+α)d(2kπ+α)=-cos(2kπ+α)+c=-cosα+c
图例解析如下：
∫cos(2kπ+α)dα=∫cos(2kπ+α)d(2kπ+α)=sina(2kπ+α)+c=sinα+c
图例解析如下：
∫tan(2kπ+α)dα=∫[sin(2kπ+α) d(2kπ+α)/ cos(2kπ+α)]=-∫d cos(2kπ+α)/cos(2kπ+α)=-ln|cos(2kπ+α)|+c=-ln|cosα|+c
图例解析如下：
∫ctg(2kπ+α)dα=∫[cos(2kπ+α) d(2kπ+α)/ sin(2kπ+α)]=∫d sin(2kπ+α)/sin(2kπ+α)=ln|sin(2kπ+α)|+c=ln|sinα|+c
图例解析如下：
∫sec(2kπ+α)dα=∫dα/ cos(2kπ+α)=∫d(2kπ+α)/ cos(2kπ+α)=∫cos(2kπ+α)d(2kπ+α)/ [cos(2kπ+α)]^2=∫dsin(2kπ+α)/ {1-[sin(2kπ+α)]^2}=∫dsin(2kπ+α)/ {[1-sin(2kπ+α)][1+ sin(2kπ+α)]}=(1/2){∫dsin(2kπ+α)/ [1-sin(2kπ+α)]+∫dsin(2kπ+α)/ [1+sin(2kπ+α)]}=(1/2)ln{[1+sin(2kπ+α)]/ [1-sin(2kπ+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：
∫csc(2kπ+α)dα=∫dα/ sin(2kπ+α)=∫d(2kπ+α)/ sin(2kπ+α)=∫sin(2kπ+α)d(2kπ+α)/ [sin(2kπ+α)]^2=-∫dcos(2kπ+α)/ {1-[cos(2kπ+α)]^2}=-∫dcos(2kπ+α)/ {[1-cos(2kπ+α)][1+ cos(2kπ+α)]}=-(1/2){∫dcos(2kπ+α)/ [1-cos(2kπ+α)]+∫dcos(2kπ+α)/ [1+cos(2kπ+α)]}=-(1/2)ln{[1+cos(2kπ+α)]/ [1-cos(2kπ+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+ctga|+c
图例解析如下：