Answer
$-\sqrt {16-y^2}+c$
Work Step by Step
Consider the integral $\int \dfrac{y dy}{\sqrt {16-y^2}}$
The given integral can be re-written as: $\int \dfrac{y dy}{\sqrt {16-y^2}}=\int (16-y^2)^{(-1/2)}y dy$
This implies that $(\dfrac{-1}{2})\int (16-y^2)^{(-1/2)}(-2y) dy=-\sqrt {16-y^2}+c$