To solve the given integral using the method you outlined, we need to evaluate:

$\int_0^{\frac{\pi}{4}} \tan^2 x \sec x \, dx$

Step 1: Rewrite $\tan^2 x$ using its identity

We can use the identity: $\tan^2 x = \sec^2 x - 1$

Thus, the integral becomes: $\int_0^{\frac{\pi}{4}} \tan^2 x \sec x \, dx = \int_0^{\frac{\pi}{4}} (\sec^2 x - 1) \sec x \, dx$

Step 2: Distribute $\sec x$

Now, multiply both terms inside the integral by $\sec x$: $\int_0^{\frac{\pi}{4}} \sec^3 x \, dx - \int_0^{\frac{\pi}{4}} \sec x \, dx$

We now have two separate integrals to solve:

1. $\int_0^{\frac{\pi}{4}} \sec^3 x \, dx$
2. $\int_0^{\frac{\pi}{4}} \sec x \, dx$

Step 3: Solve the second integral $\int_0^{\frac{\pi}{4}} \sec x \, dx$

The standard result for the integral of $\sec x$ is: $\int \sec x \, dx = \ln|\sec x + \tan x| + C$

Thus, evaluating this from 0 to $\frac{\pi}{4}$, we get: $\left[ \ln|\sec x + \tan x| \right]_0^{\frac{\pi}{4}}$

At $x = \frac{\pi}{4}$, $\sec \frac{\pi}{4} = \sqrt{2}$ and $\tan \frac{\pi}{4} = 1$: $\ln|\sec \frac{\pi}{4} + \tan \frac{\pi}{4}| = \ln(\sqrt{2} + 1)$

At $x = 0$, $\sec 0 = 1$ and $\tan 0 = 0$: $\ln|\sec 0 + \tan 0| = \ln(1) = 0$

Thus, the result for the second integral is: $\ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1)$

Step 4: Solve the first integral $\int_0^{\frac{\pi}{4}} \sec^3 x \, dx$

This integral requires a standard reduction formula for $\sec^n x$. For $\sec^3 x$, we use the following identity: $\int \sec^3 x \, dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| + C$

Evaluating this from 0 to $\frac{\pi}{4}$, we get: $\left[ \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| \right]_0^{\frac{\pi}{4}}$

At $x = \frac{\pi}{4}$, $\sec \frac{\pi}{4} = \sqrt{2}$ and $\tan \frac{\pi}{4} = 1$, so: $\frac{1}{2} \sec \frac{\pi}{4} \tan \frac{\pi}{4} = \frac{1}{2} \cdot \sqrt{2} \cdot 1 = \frac{\sqrt{2}}{2}$ And: $\frac{1}{2} \ln|\sec \frac{\pi}{4} + \tan \frac{\pi}{4}| = \frac{1}{2} \ln(\sqrt{2} + 1)$

At $x = 0$, $\sec 0 = 1$ and $\tan 0 = 0$, so: $\frac{1}{2} \sec 0 \tan 0 = 0 \quad \text{and} \quad \frac{1}{2} \ln|\sec 0 + \tan 0| = \frac{1}{2} \ln(1) = 0$

Thus, the result for the first integral is: $\frac{\sqrt{2}}{2} + \frac{1}{2} \ln(\sqrt{2} + 1)$

Step 5: Combine the results

Now, combining the results of both integrals, we get: $\left( \frac{\sqrt{2}}{2} + \frac{1}{2} \ln(\sqrt{2} + 1) \right) - \ln(\sqrt{2} + 1)$

Simplifying: $= \frac{\sqrt{2}}{2} + \frac{1}{2} \ln(\sqrt{2} + 1) - \ln(\sqrt{2} + 1)$ $= \frac{\sqrt{2}}{2} - \frac{1}{2} \ln(\sqrt{2} + 1)$

Step 6: Final Answer

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

This corresponds to the third option, which is marked in the image.

Would you like further details or have any additional questions?

Here are 5 related questions for further practice:

1. How would you derive the reduction formula for $\sec^n x$?
2. What is the integral of $\sec^4 x$ from 0 to $\frac{\pi}{4}$?
3. How do you handle integrals involving $\tan^n x$ for $n > 2$?
4. Can you solve $\int_0^{\frac{\pi}{4}} \sec^2 x \, dx$?
5. What are other methods to solve $\int_0^{\frac{\pi}{4}} \tan^2 x \sec x \, dx$?

Tip: Always check for identities or reduction formulas when working with trigonometric integrals.