How do you evaluate #int x/sqrt(1-x^2)# from #[-1/2, 0]#?

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Answer 1 · 2016-12-13T14:01:42

The answer is #=(sqrt3-2)/2=-0.134#

#intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

Let #u=1-x^2#, #=>#, #du=-2xdx#

#int(xdx)/sqrt(1-x^2)=-1/2int(du)/sqrtu#

#=-1/2intu^(-1/2)du=-1/2u^(1/2)/(1/2)#

#=-sqrt(1-x^2)#

So,

#int_(-1/2)^0(xdx)/sqrt(1-x^2)=[-sqrt(1-x^2)]_(-1/2)^0#

#=-1+sqrt(1-1/4)=-1+(sqrt3/2)#

#=(sqrt3-2)/2=-0.134#