Question

∫（pai／2到0）sinx／8+sin^2x dx

Answer 1

∫（π／2到0）[sinx／(8+sin²x)] dx=-∫(0,π/2)dcosx/(9-cos²x)=-∫(0,π/2)dcosx/(3+cosx)(3-cosx)=(-1/6)∫(0,π/2)[1/(3+cosx)+1/(3-cosx)]dcosx=(-1/6)[ln(3+cosx)-ln(3-cosx)]|(0,π/2)=(-1/6)×[(...