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

Question

2372 views around the world

You can reuse this answer
Creative Commons License

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,

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

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

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

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