根号下x平方-a平方dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 11:19:28
是这个积分么?
I=∫根号下(X^2-A^2)dX(A>0)=X根号下(X^2-A^2)--∫(X^2)dX/[根号下(X^2-A^2)][分部积分]=X根号下(X^2-A^2)--∫(X^2-A^2)dX/[根号下
设x=sint,dx=costdt,(以下省略积分符号)原式=[(sint)^2/cost]costdt=(sint)^2dt=(1-cos2t)/2*dt=1/2[dt-cos2tdt)=1/2t-
点击放大:
[x√(1-y²)]dx+[y√(1-x²)]dy=0[y√(1-x²)]dy=-[x√(1-y²)]dx分离变量得ydy/√(1-y²)=-xdx/
∫dx/[x^2√(x^2+a^2)]=∫dx/[x^3√(1+(a/x)^2)]=∫-1/2*d(1/x^2)*[1/√(1+(a/x)^2)]=∫-1/(2a^2)*d(a/x)^2*[1/√(1
令x=asin(t)就做出来了...答案是-根号下a平方-x平方再问:能详细写下积分过程吗?谢谢。再答:换元积分,微积分里有的~
x(1-y^2)^(1/2)dx+y(1-x^2)^(1/2)dy=0,|x|
∫dx/[x^2√(1+x^2)]换元,x=tant=∫d(tant)/[tan^2t√(1+tan^2)]=∫(dt/cos^2t)/[tan^2t/cost]=∫dt/cost*tan^2t=∫c
由∫10²/√(1-x²)dx令t=cosx,dx=-sinxdt∴∫10²/√(1-cos²t)(-sintdt)=-∫100dt=-100t+C=-100a
令x=a•tanθ,dx=a•sec²θdθ∫(x²+a²)^(3/2)dx=∫(a²•tan²θ+a²
这个题要用换元积分法,是令x=asect在慢慢往后算
令x=asinu,√(a²-x²)=acosu,则dx=acosudu原式=∫a²cos²udu=a²/2∫(1+cos2u)du=a²/2
令x=2sint则dx=2costdt原式=∫2cost*2costdt=2∫(1+cos2t)dt=2[t+0.5sin2t]+C=2t+sin2t+C=2arcsin(x/2)+2*x/2*√(1
∫1/[x√(1-x²)]dx=∫1/[x*√[x²(1/x²-1)]dx=∫1/[x*|x|*√(1/x²-1)]dx=∫1/[x²√(1/x