Answer
$\dfrac{1}{2}(\tan^{-1} \dfrac{x}{5}- \dfrac{5x}{x^2+25})$
Work Step by Step
$\sec{\theta} =\sqrt{1+\tan^2{\theta}} = \dfrac{1}{5} \sqrt{25+x^2}$
$\cos{\theta} = \dfrac{5}{\sqrt{25+x^2}}$
$\sin{\theta} =\dfrac{ \tan{\theta} }{\sec{\theta}} = \dfrac{x}{\sqrt{25+x^2}}$
$\sin{2\theta} =2 \sin{\theta} \cos{\theta} =2 \dfrac{x}{\sqrt{25+x^2}} \dfrac{5}{\sqrt{25+x^2}} = \dfrac{10x}{x^2+25}$
$\dfrac{\theta}{2} - \dfrac{\sin{2\theta}}{4} = \dfrac{1}{2}(\tan^{-1} \dfrac{x}{5}- \dfrac{5x}{x^2+25})$
