= Solution
First interpret the <surface integral> through the given <parametric surface>, counting multiplicity. Set
$$
J(u)=\frac{\sin u}{\sqrt{1+\sin^2u}}.
$$
Its <derivative> is
$$
J'(u)=\frac{\cos u}{\sqrt{1+\sin^2u}}-\frac{\sin^2u\cos u}{(1+\sin^2u)^{3/2}}
=\frac{\cos u}{(1+\sin^2u)^{3/2}}.
$$
Using part (c), the <fundamental theorem of calculus> gives
$$
\int_c^d\int_a^b\kappa(\mathbf r(u,v))\sqrt{EG-F^2}\,du\,dv
=(d-c)\bigl(J(b)-J(a)\bigr).
$$
On the other hand part (d), including its <arc length> factor, gives
$$
\int_{\gamma_a([c,d])}\kappa_g\,ds
=-\int_c^d\frac{\sin a}{\sqrt{1+\sin^2a}}\,dv=-(d-c)J(a),
$$
and the corresponding expression for $b$ is $-(d-c)J(b)$. Subtraction proves
$$
\boxed{\int_c^d\int_a^b\kappa(\mathbf r(u,v))\sqrt{EG-F^2}\,du\,dv
=\int_{\gamma_a([c,d])}\kappa_g\,ds-\int_{\gamma_b([c,d])}\kappa_g\,ds.}
$$
If $0<b-a<2\pi$, the <parametric surface> is <injective> on the rectangle, so this is exactly the stated <surface integral> over its image. It also remains the ordinary image integral at $b-a=2\pi$, since the repeated seam has zero <surface area>.
\b[The assertion for arbitrary $a<b$ needs a multiplicity convention or the restriction $b-a\leq2\pi$.] This omission is present in the native PDF. For example, take $a=0$, $b=5\pi/2$, $c=0$, $d=1$. The image is the entire cylindrical band $0\leq z\leq1$, since $u$ already covers a full turn. Its ordinary <surface integral> of <Gaussian curvature> is
$$
\int_0^1\int_0^{2\pi}\frac{\cos u}{(1+\sin^2u)^{3/2}}\,du\,dv=J(2\pi)-J(0)=0.
$$
But the difference of the two signed <geodesic curvature> integrals is $J(5\pi/2)-J(0)=1/\sqrt2$. Thus ordinary integration over the image set makes the unrestricted assertion false. The boxed parameter integral is valid for all $a<b$ and $c<d$ and is the appropriate corrected statement. This distinction is the <multiplicity of a parametrized surface integral>.
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