Trigonometry and Calculus!

In denominator of $F(x)$ use $\small{\color{teal}{\cos^2 x=1-\sin^2 x}}$ and than take $\sin x\cos x$ in the denominator so that:

$F(x)=\dfrac{1}{\sqrt{\dfrac{9\sin^2 x+16}{\sin^2x\cos^2 x}}}=\dfrac 1{\sqrt{9\sec^2 x+16\sec^2 x\csc^2 x}}$

Then use $\small{\color{teal}{\sec^2 x=1+\tan^2x}}$ and $\small{\color{teal}{\csc^2 x=1+\cot^2x}}$ so that:

$F(x)=\dfrac{1}{\sqrt{41+25\tan^2 x+16\cot^2 x}}$

Note by $AM\ge GM$ , $\small{25\tan^2 x+16\cot^2 x\ge 2\sqrt{25\cdot 16\cdot\underbrace{ \tan^2x \cdot\cot^2x}_1}=40}$ So denominator is minimum (and hence $F(x)$ is maximum) when $\small{25\tan^2 x=16\cot^2 x}$ or $\small{\tan^4 x=\frac{16}{25}\text{ or }x=\tan^{-1}\left(\frac 2{\sqrt 5}\right)=\sin^{-1}\left(\frac 2{ 3}\right)}$ ,

$\left(F(x)\right)_{\text{max}}=\dfrac{1}{\sqrt{41+40}}=\dfrac 19~\text{at}~x=\sin^{-1}\left(\frac 2{ 3}\right)$ .Now

$\mathcal H=\displaystyle\int_{0}^{\sin^{-1}\left(\frac 2{ 3}\right)}\dfrac{\sin x\cos x}{\sqrt{9\sin^2 x+16}}\mathrm{d}x$

Substitute $9\sin^2 x+16=u$ such that $18\sin x\cos x \mathrm{d}x=\mathrm{d}u$ so that:

$\mathcal H=\dfrac 1{18}\displaystyle\int_{16}^{20 }\dfrac{\mathrm{d}u}{\sqrt{u}}$

$=\dfrac{1}{18}\left[2\sqrt u\right]_{16}^{20}$

$\Large =\dfrac{2}9\left(\sqrt 5-2\right)$

Hence, $\Large 2+9+5+2=\boxed{18}$ .