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)
