Question

∫dx/(sin²x+tan²x)

Answer 1

问：∫dx/(sin²x+tan²x)

答：您好，很高兴回答您的问题。∫dx/(sin²x+tan²x)解答：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, 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