∫dx/(sin²x+tan²x) 求过程

Question

如题所述

Answer 1

符号难敲
∫dx/(sin²x+tan²x)
=∫[cos²x/(sin²xcos²+sin²x)]dx
=∫[(1+cos2x)/2]/[(1/4)sin²2x+(1-cos2x)/2]dx
=(1/2)∫[(1+cos2x)]/[(1/4)(1-cos²2x+(1-cos2x)/2]dx
=-2∫[(1+cos2x)]/[(cos²2x+2cos2x-3]dx
=(1/2)∫[1/(cost+3)]dt+(1/2)∫[1/(cost-1)]dt
对于∫[1/(cosx+3)]dx这类积分，可以参考：
万能代换t＝tan(x/2)，则x＝2arctant，dx＝2dt/(1+t^2)，cosx＝(1-t^2)/(1+t^2)，所以
∫dx/(cosx+3)＝∫dt/(t^2+2)＝1/√2×arctan(t/√2)＋C＝1/√2×arctan(tan(x/2)/√2)＋C本回答被提问者和网友采纳