## University Calculus: Early Transcendentals (3rd Edition)

$\sin^{-1} (\dfrac{\theta}{4})+C$
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, $= \sin^{-1} (\dfrac{\theta}{4})+C$