Question

∫dx/(sin²x+tan²x)

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,
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

Answer 2

楼上万能代换太繁琐了。。不如上下同乘sec²x，分母变为tan²x+sec²xtan²x，进一步转化为2tan²x+tan⁴x然后凑dtanx，换元成u转化成有理分式的积分即可。