#cot(arcsinx) = (cos(arcsinx))/(sin(arcsinx))#
#= (cos(arcsinx))/x#
Normally, here would be enough for a high school Pre-Calculus problem. But, let's try something creative.
#d/(dx)[cos(arcsinx)] = -sin(arcsinx) * 1/(sqrt(1-x^2)) = -x/sqrt(1-x^2)#
Let:
#u = 1-x^2#
#du = -2xdx#
#int (-x)/(sqrt(1-x^2))dx = 1/2int (-2x)/(sqrt(1-x^2))dx#
#= 1/2int 1/(sqrt(u))du#
#= 1/2int u^(-1/2)du#
#= 1/2*[2sqrtu] = sqrtu = sqrt(1-x^2)#
There, now we have a representation for #cos(arcsinx)#.
#int{d/(dx)[cos(arcsinx)]}dx = cos(arcsinx)#
We haven't changed the domain, since the domain of #sinx# and #cosx# are larger than that of #arcsinx#. Thus, #arcsinx# decides the domain restrictions on the left and right boundaries. Remember the #x# in the denominator adds a vertical asymptote at #x = 0#.
#=> (cos(arcsinx))/x = sqrt(1-x^2)/x#
#AA x in [-1, 0) uu (0, 1]#
(For all #x# in the domain of #-1 <= x < 0# and #0 < x <= 1#)
