dx/cos(a+x)cos(b+x)求不定积分
dx/cos(a+x)cos(b+x)求不定积分
如题
如题
其他人气:703 ℃时间:2020-01-26 06:12:54
优质解答
cosacosb=(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)/2dx=1/1+t^2cos(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...设a^2=[1+cos(a-b)]/[1-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)这部分看不懂,a不是常数吗,怎么可以设a^2=[1+cos(a-b)]/[1-cos(a-b)] >0这里设a,b,是常数,所以cos (a-b)也是常数 1+cos(a-b)]/[1-cos(a-b)>0所以设正a^2=1+cos(a-b)]/[1-cos(a-b);,后边要因式分解,写起来略微简单点
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