Question

定积分∫[0,π](tanxsecx)/(sec²x+1)dx怎么算

这一题用牛顿莱布尼茨需不需要变成[0,π/2]和[π/2,π]两个区间?
其实我最主要想问的是∫[0,π](tanxsecx)/(sec²x+1)dx和∫[0,π](sinx)/(cos²x+1)dx这两个是同一个积分么?后者上下同除以cos²x就变成了前者，但是我算出来两个答案不一样，是我算错了还是本来就是不同的积分?

Answer 1

∫[0,a]√(x2+a2) dx,令x=a*tany => dx=a*sec2y dy 当x=0,y=0 // 当x=a,y=π/4 原式= ∫[0,π/4]√(a2*tan2y+a2) * a*sec2y dy = ∫[0,π/4]√[a2(1+tan2y)] * a*sec2y dy = ∫[0,π/4]a*secy*a*sec2y dy = a2∫[0,π/4]sec3y dy = a2 * [(1/2)secy*tany + (1/2)ln|secy+tany|] [0,π/4] = a2 * [(1/2)sec(π/4)tan(π/4) + (1/2)ln(sec(π/4)+tan(π/4))] - a2 * [(1/2)ln(1)] = a2 * [(1/2)(√2)(1) + (1/2)ln(√2+1)] = a2/√2 + (a2/2)ln(√2+1) 有关∫sec3x dx的积分： J = ∫sec3x dx = ∫secx dtanx = secx*tanx - ∫tanx dsecx = secx*tanx - ∫tanx*secxtanx dx = secx*tanx - ∫(sec2x-1)*secx dx = secx*tanx - J + ∫secx dx 2J = secx*tanx + ∫secx(secx+tanx)/(secx+tanx) dx J = (1/2)secx*tanx + (1/2)∫(secxtanx+sec2x)/(secx+tanx) dx J = (1/2)secx*tanx + (1/2)∫d(secx+tanx)/(secx+tanx) dx J = (1/2)secx*tanx + (1/2)ln|secx+tanx| + C

追问

这是啥玩意