第1个回答 2022-10-27
cosaco *** =(cos(a+b)+cos(a-b))/2
=∫2dx/[cos(a-b)+cos(2x+a+b)]
设t=tan[x+(a+b)/2]
x=atant-(a+b)/2
dx=1/1+t^2
cos(2x+a+b)=(1-t^2)/(1+t^2)
代入后
=∫2dt/{(1+cos(a-b))+(cos(a-b)-1)*t^2}
cos(a-b)=1时候
=t=tan[x+(a+b)/2+C
cos(a-b)=-1
=1/t=ctan[x+(a+b)/2]+C
否则|cos(a-b)|0
=1/[cos(a-b)-1]∫2dt/t^2-a^2
=1/{a*[cos(a-b)-1]} * ln(t-a)/(t+a)