Answer
$\sin^{-1} (\dfrac{\theta}{4})+C$
Work Step by Step
Calculate the anti-derivative.
Then $\int \dfrac{d \theta}{\sqrt {16-\theta^2}}=\int \dfrac{d \theta}{\sqrt {4^2-\theta^2}}$
We know that $ \int \dfrac{dx}{\sqrt {a^2-x^2}} dx=\sin^{-1} (\dfrac{x}{a})+C$; where $C$ is a constant of proportionality.
Thus, $= \sin^{-1} (\dfrac{\theta}{4})+C$
