第1个回答 2020-09-17
∫√（a^2-x^2）dx
设x=asint
则dx=dasint=acostdt
a^2-x^2
=a^2-a^2sint^2
=a^2cost^2
∫√（a^2-x^2）dx
=∫acost*acostdt
=a^2∫cost^2dt
=a^2∫（cos2t+1）/2dt
=a^2/4∫(cos2t+1)d2t
=a^2/4*(sin2t+2t)
将x=asint代回
∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C
扩展资料
不定积分的公式
1、∫ a dx = ax + C，a和C都是常数
2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1
3、∫ 1/x dx = ln|x| + C
4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1
5、∫ e^x dx = e^x + C
6、∫ cosx dx = sinx + C
7、∫ sinx dx = - cosx + C
第2个回答 2011-12-07
1.x/2√(x²+a²)+a²/2ln(x+√(x²+a²))+c
2.ln(x+√(x²+a²))+c本回答被提问者采纳
2.ln(x+√(x²+a²))+c本回答被提问者采纳
第3个回答 2011-12-07
1
∫√(x^2+a^2)dx [√x^2+a^2) ]'=2x*(1/2)*（1/√(x^2+a^2))=x/√(x^2+a^2)
=x√(x^+a^2)-∫xd√(x^2+a^2)
=x√(x^2+a^2)-∫x^2dx/√(x^2+a^2)
=x√(x^2+a^2)-∫√(x^2+a^2)dx+∫a^2dx/√(x^2+a^2)
2∫√(x^2+a^2dx=x√(x^2+a^2)+a^2∫d(x/a)/√[(x/a)^2+1)]
x/a=tanu d(x/a)=dtanu cosu=1/√(1+tanu^2)=a/√(x^2+a^2)
sinu=x/√(x^2+a^2)
∫d(x/a)/√[(x/a)^2+1]=∫secudu=∫dsinu/[(1-sinu)(1+sinu)]
=(1/2)ln|(1+sinu)/(1-sinu)|
=ln|(1+sinu)/cosu|
=ln[|x+√(x^2+a^2)|/|a|]
∫√x^2+a^2)dx=(1/2)x√(x^2+a^2)+(a^2/2)ln[|x+√(x^2+a^2)|/|a|] +C
2
∫dx/√(x^2+a^2)=∫d(x/a)/√[(x/a)^2+1]=ln[|x+√(x^2+a^2)|/|a|] +C
∫√(x^2+a^2)dx [√x^2+a^2) ]'=2x*(1/2)*（1/√(x^2+a^2))=x/√(x^2+a^2)
=x√(x^+a^2)-∫xd√(x^2+a^2)
=x√(x^2+a^2)-∫x^2dx/√(x^2+a^2)
=x√(x^2+a^2)-∫√(x^2+a^2)dx+∫a^2dx/√(x^2+a^2)
2∫√(x^2+a^2dx=x√(x^2+a^2)+a^2∫d(x/a)/√[(x/a)^2+1)]
x/a=tanu d(x/a)=dtanu cosu=1/√(1+tanu^2)=a/√(x^2+a^2)
sinu=x/√(x^2+a^2)
∫d(x/a)/√[(x/a)^2+1]=∫secudu=∫dsinu/[(1-sinu)(1+sinu)]
=(1/2)ln|(1+sinu)/(1-sinu)|
=ln|(1+sinu)/cosu|
=ln[|x+√(x^2+a^2)|/|a|]
∫√x^2+a^2)dx=(1/2)x√(x^2+a^2)+(a^2/2)ln[|x+√(x^2+a^2)|/|a|] +C
2
∫dx/√(x^2+a^2)=∫d(x/a)/√[(x/a)^2+1]=ln[|x+√(x^2+a^2)|/|a|] +C
相似回答
