Begin by making a #u#-substitution.
Let #u=1-x^2#
#intxsqrt(u)dx=intu^(1/2)xdx#
#du=-2xdx#
#(du)/(-2)=(-2xdx)/(-2)#
#(-1)/2*du=xdx#
#intu^(1/2)xdx#, Note that #xdx# can be replaced with #(-1)/2*du#
#intu^(1/2)*(-1)/2*du#
#=(-1)/2intu^(1/2)du#
#=(-1)/2[u^(3/2)/(3/2)]#
#=(-1)/2[u^(3/2)*(2/3)]#
#=(-1)/2[(2u^(3/2))/3]#
#=(-1)[(u^(3/2))/3]#
#=-[(u^(3/2))/3]#, now switch from #u# back to #1-x^2#
#=-[((1-x^2)^(3/2))/3]_0^1#, remember to put the original boundaries
#=-[((1-(1)^2)^(3/2))/3-((1-(0)^2)^(3/2))/3]#
#=-[((1-1)^(3/2))/3-((1-0)^(3/2))/3]#
#=-[((0)^(3/2))/3-((1)^(3/2))/3]#
#=-[(0)/3-(1)/3]#
#=-[-1/3]#
#=1/3 -> Solution#