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$\begin{aligned} I & = \int_0^\frac \pi 4 \frac {dx}{\cos x} \\ & = \int_0^\frac \pi 4 \sec x \ dx \\ & = \int_0^\frac \pi 4 \frac {\sec x (\tan x + \sec x)}{\tan x + \sec x} \ dx \\ & = \int_0^\frac \pi 4 \frac {\sec x \tan x + \sec^2 x}{\color{#3D99F6}{\tan x + \sec x}} \ dx & \small \color{#3D99F6}{\text{Let }u=\tan x + \sec x \implies du = (\sec^2 x + \tan x \sec x) \ dx} \\ & = \int_1^{1+\sqrt 2} \frac {du}u \\ & = \ln u \ \big|_1^{1+\sqrt 2} \\ & = \ln (1+\sqrt 2) \end{aligned}$

$\begin{aligned} \implies \sinh \left( \int_0^\frac \pi 4 \frac {dx}{\cos x} \right) & = \sinh (\ln (1+\sqrt 2)) \\ & = \frac 12 \left( e^{\ln(1+\sqrt 2)} - e^{-\ln (1+\sqrt 2)} \right) \\ & = \frac 12 \left( 1+\sqrt 2 - \frac 1{1+\sqrt 2} \right) \\ & = \frac 12 \left( 1+\sqrt 2 - \sqrt 2 + 1 \right) \\ & = \boxed{1} \end{aligned}$