根号下4 64x平方dx-搜答案-搜狗做题网

xdx/(1-x*x)^(1/2)=-1/2*d(1-x*x)/(1-x*x)^(1/2)再问：我也是这样算的最后是负一但答案是1

是这个积分么?

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-

原式=1/2∫√(x²-3)dx²=1/2∫(x²-3)^1/2d(x²-3)=1/2*(x²-3)^(1/2+1)/(1/2+1)+C=1/3*(x

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/

∫1/[x√(1-ln²x)]dx=∫1/√(1-ln²x)d(lnx)=arcsin(lnx)+C公式：∫dx/√(a²-x²)=arcsin(x/a)+C

令x＝siny,则：√（1－x^2）＝√［1－（siny）^2］＝cosy,　y＝arcsinx,　dx＝cosydy.原式＝∫［cosy/（siny＋cosy）］dy　　＝∫｛cosy（cosy－s

x(1-y^2)^(1/2)dx+y(1-x^2)^(1/2)dy=0,|x|

∫(-2→-1)√(3-4x-x^2)dx=∫(-2→-1)√[7-(x+2)^2]dxx+2=√7sinθ、dx=√7cosθdθθ∈[0,arcsin(1/√7)]=∫(√7cosθ)(√7cos

由∫10²/√（1-x²）dx令t=cosx,dx=-sinxdt∴∫10²/√（1-cos²t）（-sintdt）=-∫100dt=-100t+C=-100a

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