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can you solve this by turning tan^2x to its sec identity, then after multiplying the sec identity with the secx, separate the integrals and solve them, then plug the limit back in 

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.

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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.

Can you solve this by turning tan^2x to its sec identity, then after multiplying the sec identity with sec x, separate the integrals and solve them, then plug the limit back in?

This tutorial covers how to evaluate the integral ∫_0^π/4 tan^2(x) sec(x) dx by transforming tan^2(x) into secant identities, separating integrals, and applying reduction formulas. Perfect for students learning advanced calculus techniques involving trigonometric integrals.

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