Trig integral

Integrating by parts, and using the substitution $u = \tan x$ gives $\begin{array}{rcl} \displaystyle \int_0^{\frac12\pi} \frac{14x \sin x \cos x}{(16\cos^2x + 9\sin^2x)^2}\,dx & = & \displaystyle \left[ \frac{x}{16\cos^2x + 9\sin^2x}\right]_0^{\frac12\pi} - \int_0^{\frac12\pi} \frac{dx}{16\cos^2x + 9\sin^2x} \\ & = & \displaystyle \frac{\pi}{18} - \int_0^{\frac12\pi} \frac{\sec^2x\,dx}{16 + 9\tan^2x} \\ & = & \displaystyle \frac{\pi}{18} - \int_0^\infty \frac{du}{16 + 9u^2} \; = \; \frac{\pi}{18} - \Big[ \frac{1}{12}\tan^{-1}\big(\tfrac{3u}{4}\big)\Big]_0^\infty \\ & = & \displaystyle \frac{\pi}{18} - \frac{\pi}{24} \; = \; \frac{\pi}{72} \end{array}$ so that $\int_0^{\frac12\pi} \frac{x \sin x \cos x}{(16\cos^2x + 9\sin^2x)^2}\,dx \; = \; \frac{\pi}{ 1008} \;,$ and so the answer is $1008 \div 4 \,=\, \boxed{252}$ .