定积分∫(x^2+a^2) ^ 1/2dx积分上限—a积分下限0 积分上限啊,下限0,1/(x^2+a^2)^3/2dx(a＞...
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\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u4ee4x=sin\u0398
dx=cos\u0398d\u0398
x=1/2\uff0c\u0398=\u03c0/6
x=0,\u0398=0
\u539f\u5f0f=\u222b(\u03c0/6,0)cos\u0398*cos\u0398d\u0398
=\u222b(\u03c0/6,0)(1+cos2\u0398)/2*1/2d(2\u0398)
=1/4*(sin2\u0398+2\u0398)|(\u03c0/6,0)
=\u221a3/8+\u03c0/12
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\u89e3
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
可用分部积分法，设原积分值是A，则
A＝∫(0到a)√(x^2+a^2)dx //选u=√(x^2+a^2)，v=x
＝a×√2a－∫(0到a) x^2/√(x^2+a^2)dx //将分子的x^2看作(x^2+a^2)-a^2
＝a×√2a－∫(0到a) √(x^2+a^2)dxa×√2a＋a^2×∫(0到a) 1/√(x^2+a^2)dx //利用不定积分公式，1/√(x^2+a^2)的原函数是ln|x+√(x^2+a^2)|
＝a×√2a－A＋a^2×ln(1+√2)
得A＝√2a^2/2＋a^2/2×ln(1+√2)，此为原积分值
令x=asint，0≤t≤π/2.则解得t=arcsin（x/a）
∫√（a²-x²)dx
=∫√[a²-a²sint]d(asint)
=a²∫cos²tdt=(a²/2)∫(1+cos2t)dt
=(a²/2)[t+(1/2)sin2t]+C
=(a²/2)[arcsin(x/a)+(x/a²)√(a²-x²)]+C
=(a²/2)arcsin(x/a)+(x/2)√(a²-x²)+C(C表示常数）
∫（0，a）√(a²-x²)dx=a²∫(0,π/2)cos²tdt=(a²/2)[t+(sin2t)/2](0,π/2)=πa²/4.
∫√(x^2+a^2)dx
=(a^2)∫√[(x/a)^2+1]d(x/a)
x/a=tanu d(x/a)=dtanu=secu^2du
cosu=√[1/(1+tanu^2)]=a/√(x^2+a^2)
sinu=x/√(a^2+x^2)
=a^2∫secu^3du
=a^2∫secudtanu^2
=a^2secu(tanu)^2-a^2∫(tanu)^2secudu
=a^2secu(tanu)^2-a^2∫[(secu)^3-secu]du
=a^2secu(tanu)^2-a^2∫(secu)^3+a^2∫secudu
2a^2∫(secu)^3du=a^2secu(tanu)^2+a^2*(1/2)ln|(1+sinu)/(1-sinu)|
a^2∫(secu)^3du=(a^2/2)secu(tanu)^2+(a^2/2)ln|(1+sinu)/cosu|+C
原式=(a^2/2)√[(x/a)^2+1] *(x/a)^2 +(a^2/2)ln| (√(x^2+a^2)+x)/a|+C
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