1.
第一步凑微分∫sinxdx/(sin^6x+cos^6x)＝∫[1/(sin^6x+cos^6x)]d(sin^2x) 令t＝sin^2x＝∫dt/[t^3+(1-t)^3]＝1/3∫[(t-1/2)^2＋1/12]dt＝4∫1/(2根号3t-1)^2+1)=4arctan[(2根号3)t-1]+c=4arctan[(2根号3)sin^2x-1]+c
3
第二题我再想想
搞定了
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