Answer
$\dfrac{1}{2}(\sin^{-1} \dfrac{x}{4}- \dfrac{x \sqrt{16-x^2}}{16})$
Work Step by Step
$\sin{\theta} = \dfrac{x}{4}$
$\cos{\theta} =\sqrt{1-\sin^2{\theta}} = \dfrac{1}{4} \sqrt{16-x^2}$
$\sin{2\theta} = 2 \sin{\theta} \cos{\theta} = \dfrac{x \sqrt{16-x^2}}{8}$
$\dfrac{\theta}{2} - \dfrac{\sin{2\theta}}{4} = \dfrac{1}{2}(\sin^{-1} \dfrac{x}{4}- \dfrac{x \sqrt{16-x^2}}{16})$