Answer
$csc(x)$
Work Step by Step
Step 1. Let $z=cos(x)$; we have $\frac{dz}{dx}=-sin(x)$
Step 2. Given
$y=\int_{cos(x)}^0(\frac{1}{1-t^2})dt=\int_0^{cos(x)}(-\frac{1}{1-t^2})dt=\int_0^{z}(-\frac{1}{1-t^2})dt$
Step 3. Based on the Fundamental Theorem of Calculus, we have
$\frac{dy}{dz}=(-\frac{1}{1-z^2})$
which leads to
$\frac{dy}{dx}=(-\frac{1}{1-z^2})\frac{dz}{dx}=(-\frac{1}{1-(cos(x))^2})(-sin(x))=\frac{sin(x)}{(sin(x))^2}=csc(x)$
