1. Use standard inverse-trigonometric identities:

For any real $x$, $\sec^2(\tan^{-1}x)=1+x^2$ and $\csc^2(\cot^{-1}x)=1+x^2$ because if $\theta=\tan^{-1}x$, then $\tan\theta=x$ and $\sec^2\theta=1+\tan^2\theta=1+x^2$. Similarly, if $\phi=\cot^{-1}x$, then $\cot\phi=x$ and $\csc^2\phi=1+\cot^2\phi=1+x^2.$

2. Apply these to the given equation: $\sec ^2(\tan ^{-1} \alpha)+\csc ^2(\cot ^{-1} \beta)=36$ so $1+\alpha^2+1+\beta^2=36$ $\alpha^2+\beta^2=34.$

3. Also given: $\alpha+\beta=8.$

Use $(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$ so $64=34+2\alpha\beta$ $2\alpha\beta=30$ $\alpha\beta=15.$

4. Hence $\alpha,\beta$ are roots of $t^2-8t+15=0$ which factors as $(t-3)(t-5)=0.$

Thus the two numbers are $3$ and $5$. Since $\alpha\le \beta$, we get $\alpha=3,\quad \beta=5.$

5. Now compute: $\alpha^2+\beta=3^2+5=9+5=14.$

Therefore, the required integer is $\boxed{14}.$