The integral to evaluate is:

$\int_0^{\frac{\pi}{4}} 6 \sec^3(x) \, dx$

To solve this, we'll break it down step by step. The integral of $\sec^3(x)$ is not straightforward, but we can use a standard reduction formula. The formula for the integral of $\sec^3(x)$ is:

$\int \sec^3(x) \, dx = \frac{1}{2} \sec(x) \tan(x) + \frac{1}{2} \ln |\sec(x) + \tan(x)| + C$

So, applying this to the given integral:

$6 \int_0^{\frac{\pi}{4}} \sec^3(x) \, dx = 6 \left[ \frac{1}{2} \sec(x) \tan(x) + \frac{1}{2} \ln |\sec(x) + \tan(x)| \right]_0^{\frac{\pi}{4}}$

Step 1: Evaluate at $x = \frac{\pi}{4}$

At $x = \frac{\pi}{4}$, we have: $\sec\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \tan\left(\frac{\pi}{4}\right) = 1$ Substitute into the expression: $\frac{1}{2} \sec\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) = \frac{1}{2} \times \sqrt{2} \times 1 = \frac{\sqrt{2}}{2}$ $\frac{1}{2} \ln\left|\sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right)\right| = \frac{1}{2} \ln(\sqrt{2} + 1)$

Step 2: Evaluate at $x = 0$

At $x = 0$, we have: $\sec(0) = 1, \quad \tan(0) = 0$ Substitute into the expression: $\frac{1}{2} \sec(0) \tan(0) = \frac{1}{2} \times 1 \times 0 = 0$ $\frac{1}{2} \ln\left|\sec(0) + \tan(0)\right| = \frac{1}{2} \ln(1 + 0) = \frac{1}{2} \ln(1) = 0$

Step 3: Combine the results

Now, we subtract the values at $x = 0$ from the values at $x = \frac{\pi}{4}$: $6 \left( \frac{\sqrt{2}}{2} + \frac{1}{2} \ln(\sqrt{2} + 1) - 0 \right) = 3\sqrt{2} + 3 \ln(\sqrt{2} + 1)$

Thus, the final answer is: $\boxed{3\sqrt{2} + 3 \ln(\sqrt{2} + 1)}$

Would you like any further details or clarification on the steps?

Here are 5 related questions you might find helpful:

1. How do you integrate $\sec^n(x)$ for different values of $n$?
2. What are the properties of the secant and tangent functions in integrals?
3. How can you use reduction formulas for more complex trigonometric integrals?
4. Why do logarithmic functions appear in the integration of $\sec^3(x)$?
5. How do definite integrals differ from indefinite integrals in terms of evaluation?

Tip: For integrals involving powers of secant, memorizing the reduction formulas for powers like $\sec^3(x)$ can save time!