How do you test the improper integral #int x/sqrt(1-x^2)dx# from #[0,1]# and evaluate if possible?

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Answer 1 · 2017-03-31T20:23:00

#1#

Making #x=sin y# with #dx=cosy dy# and substituting into the integral

#I=int (x dx)/sqrt(1-x^2) equiv int (sin y/cosy)cosy dy = int siny dy = -cosy#

The integration limits are changed to

#x=0 -> y = 0#
#x=1->y->pi/2#

so

#I=-cos(pi/2)-(-cos(0)) = 1#

Answer 2 · 2017-03-31T23:59:05

#int_0^1x/sqrt(1-x^2)dx=1#

We can also use the substitution #u=1-x^2# implying that #du=-2xdx#.

#intx/sqrt(1-x^2)dx=-1/2int(-2xdx)/sqrt(1-x^2)=-1/2intu^(-1/2)du#

#=-1/2(u^(1/2)/(1/2))=-sqrtu=-sqrt(1-x^2)+C#

So the integral with bounds is:

#int_0^1x/sqrt(1-x^2)dx=[-sqrt(1-x^2)]_0^1=-sqrt(1-1)-(-sqrt(1-0))=1#