Question

定积分∫(x^2+a^2) ^ 1/2dx积分上限—a积分下限0

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