∫dx/[(2+cosx)sinx]
=∫sinxdx/[(2+cosx)sin²x]
=∫-d(cosx)/[(2+cosx)(1-cos²x)]
=∫{1/[(cosx+1)(cosx-1)]-1/[(cosx-1)(cosx+2)]}d(cosx)
=∫{1/[2(cosx-1)]-1/[2(cosx+1)-1/[3(cosx-1)]+1/[3(cosx+2)]}d(cosx)
=(1/2)∫d(cosx)/(cosx-1)-(1/2)∫d(cosx)/(cosx+1)-(1/3)∫d(cosx)/(cosx-1)+(1/3)∫d(cosx)/(cosx+2)
=(1/2)ln|cosx-1|-(1/2)ln|cosx+1|-(1/3)ln|cosx-1|+(1/3)ln|cosx+2|
=(1/6)ln|cosx-1|+(1/3)ln|cosx+2|-(1/2)ln|cosx+1|
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