Let’s break down and solve the double integral step-by-step: \[ \int_{y=0}^{1} \left( \int_{x=\arctan(y)}^{\frac{\pi}{4}} \sec(x) \, dx \right) dy \] --- ### **Step 1: Integrate with respect to \( x \)** The inner integral is: \[ \int_{\arctan(y)}^{\frac{\pi}{4}} \sec(x) \, dx \] Recall: \[ \int \sec(x)\,dx = \ln |\sec(x) + \tan(x)| + C \] So, \[ \int_{\arctan(y)}^{\frac{\pi}{4}} \sec(x) \, dx = \left[\ln |\sec(x) + \tan(x)| \right]_{\arctan(y)}^{\frac{\pi}{4}} \] \[ = \ln |\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln |\sec(\arctan(y)) + \tan(\arctan(y))| \] --- ### **Step 2: Simplify the bounds** - \(\sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos(\frac{\pi}{4})} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}\) - \(\tan\left(\frac{\pi}{4}\right) = 1\) Thus, \[ \sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4}) = \sqrt{2} + 1 \] For \(x = \arctan(y)\): - \(\tan(\arctan(y)) = y\) - \(\sec(\arctan(y)) = \frac{1}{\cos(\arctan(y))}\) Recall: If \(\tan(\theta) = y\), then \(\sin(\theta) = \frac{y}{\sqrt{1 + y^2}}\), \(\cos(\theta) = \frac{1}{\sqrt{1 + y^2}}\), so \(\sec(\theta) = \sqrt{1 + y^2}\). Thus, \[ \sec(\arctan(y)) + \tan(\arctan(y)) = \sqrt{1 + y^2} + y \] --- ### **Step 3: Plug into the outer integral** So the expression to integrate with respect to \(y\) is: \[ \ln(\sqrt{2} + 1) - \ln(\sqrt{1+y^2} + y) = \ln\left(\frac{\sqrt{2}+1}{\sqrt{1+y^2}+y}\right) \] So the outer integral is: \[ \int_0^1 \ln\left(\frac{\sqrt{2}+1}{\sqrt{1+y^2}+y}\right) dy \] \[ = \int_0^1 \left[\ln(\sqrt{2}+1) - \ln(\sqrt{1+y^2}+y)\right] dy \] \[ = \ln(\sqrt{2}+1) \int_0^1 dy - \int_0^1 \ln(\sqrt{1+y^2}+y) dy \] \[ = \ln(\sqrt{2}+1) [y]_0^1 - \int_0^1 \ln(\sqrt{1+y^2}+y) dy \] \[ = \ln(\sqrt{2}+1) (1-0) - \int_0^1 \ln(\sqrt{1+y^2}+y) dy \] \[ = \ln(\sqrt{2}+1) - \int_0^1 \ln(\sqrt{1+y^2}+y) dy \] --- ### **Step 4: Evaluate the remaining integral** Let’s try to simplify or compute \(\int_0^1 \ln(\sqrt{1+y^2} + y) dy\). Let \(y = \sinh t\). Then \(dy = \cosh t dt\), \(y = 0 \rightarrow t = 0\), \(y=1 \rightarrow t = \sinh^{-1}(1) = \ln(1+\sqrt{2})\). \[ \sqrt{1+y^2} + y = \sqrt{1+\sinh^2 t} + \sinh t = \cosh t + \sinh t = e^t \] So: \[ \int_0^1 \ln(\sqrt{1+y^2}+y) dy = \int_{t=0}^{t=\ln(1+\sqrt{2})} \ln(e^t) \cosh t \, dt \] \[ = \int_{0}^{\ln(1+\sqrt{2})} t \cosh t \, dt \] This can be integrated by parts: Let \(u = t\), \(dv = \cosh t dt\) Then \(du = dt\), \(v = \sinh t\) So, \[ \int t \cosh t \, dt = t \sinh t - \int \sinh t \, dt = t \sinh t - \cosh t + C \] So, \[ \int_{0}^{\ln(1+\sqrt{2})} t \cosh t \, dt = \left[ t \sinh t - \cosh t \right]_{0}^{\ln(1+\sqrt{2})} \] --- ### **Step 5: Plug in the values** At \(t = \ln(1+\sqrt{2})\): - \(\sinh t = \sinh(\ln(1+\sqrt{2})) = 1\) (since \(\sinh(\ln(x + \sqrt{x^2+1})) = x\), for \(x=1\)) - \(\cosh t = \sqrt{1 + \sinh^2 t} = \sqrt{1+1^2} = \sqrt{2}\) At \(t=0\): - \(\sinh 0 = 0\) - \(\cosh 0 = 1\) So, \[ t \sinh t - \cosh t \Big|_{0}^{\ln(1+\sqrt{2})} = [\ln(1+\sqrt{2}) \cdot 1 - \sqrt{2}] - [0 - 1] = \ln(1+\sqrt{2}) - \sqrt{2} + 1 \] --- ### **Step 6: Write the final answer** Recall the earlier result: \[ \boxed{ \ln(\sqrt{2}+1) - \left[ \ln(1+\sqrt{2}) - \sqrt{2} + 1 \right] } \] \[ = \ln(\sqrt{2}+1) - \ln(1+\sqrt{2}) + \sqrt{2} - 1 \] \[ = \sqrt{2} - 1 \] **Final Answer:** \[ \boxed{\sqrt{2} - 1} \]

The value of the given double integral is: \[ \boxed{\sqrt{2} - 1} \]