解ï¼1=(sinx)^2+(cosx)^2, 被积åå½æ°å¯ä»¥å解为:
1/sinx(1+cosx)+1/(1+cosx); 1/(1+cosx)=(1/2)[cos(x/2)]^2........(1); ä¸é¢å©ç¨ä¸è½å
¬å¼ï¼
sinx=2tan(x/2)/[1+tan^2(x/2)],
(1+cosx)=1+[1-tan^2(x/2)]/[1+tan^2(x/2)]=2/[1+tan^2(x/2)].
1/[sinx(1+cosx)]=(1/4)[1+tan^2(x/2)]^2/tan(x/2)
=(1/4)[cot(x/2)+2tan(x/2)+tan^3(x/2)]=(1/4)[cot(x/2)+tan(x/2)+sin(x/2)/cos^3(x/2)]..............................(2);
å 为ï¼(1/4)tan^3(x/2)=(1/4)sin(x/2)[1-cos^2(x/2)]/cos^3(x/2)
=(1/4)sin(x/2)/cos^3(x/2)-(1/4)tan(x/2)ï¼è¢«ç§¯åå¼=(1)+(2)
åå¼=â«d[tan(x/2)+(1/2)â«dsin(x/2)/sin(x/2)-(1/2)â«[dcos(x/2) /cos^3(x/2)]
=tan(x/2)+(1/2)[ln|sin(x/2)|-ln|cos(x/2)|]-(1/2)*(-1/2)/cos^2(x/2)+C
=tan(x/2)+1/[4cos^2(x/2)]+(1/2)ln|tan(x/2)|+C
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