To solve the given integral:

$\int_{0}^{\pi} \frac{1}{2} \frac{1}{\cos x + |\cos x|} \, dx$

Step 1: Simplify the term $\cos x + |\cos x|$

The expression $\cos x + |\cos x|$ depends on the sign of $\cos x$:

• When $\cos x \geq 0$, $|\cos x| = \cos x$, so $\cos x + |\cos x| = \cos x + \cos x = 2\cos x$.
• When $\cos x < 0$, $|\cos x| = -\cos x$, so $\cos x + |\cos x| = \cos x - \cos x = 0$.

This means:

• For $\cos x \geq 0$, the integrand becomes $\frac{1}{4\cos x}$.
• For $\cos x < 0$, the integrand is undefined since division by 0 is not possible.

Step 2: Determine the intervals

The cosine function $\cos x$ is non-negative on $[0, \pi/2]$ and negative on $(\pi/2, \pi]$. Therefore:

1. On $[0, \pi/2]$, the integral is valid.
2. On $(\pi/2, \pi]$, the integral is not defined.

Step 3: Compute the integral on $[0, \pi/2]$

The integral reduces to: $\int_{0}^{\pi/2} \frac{1}{4\cos x} \, dx$ This is equivalent to: $\frac{1}{4} \int_{0}^{\pi/2} \frac{1}{\cos x} \, dx$ The integral of $\frac{1}{\cos x}$ is $\ln|\sec x + \tan x|$. Evaluating from $0$ to $\pi/2$, we have: $\frac{1}{4} \left[ \ln|\sec x + \tan x| \right]_{0}^{\pi/2}$

Step 4: Evaluate the limits

• At $x = 0$: $\sec 0 = 1, \quad \tan 0 = 0 \quad \implies \ln|\sec 0 + \tan 0| = \ln(1) = 0.$
• At $x = \pi/2$: $\sec(\pi/2) \to \infty, \quad \tan(\pi/2) \to \infty \quad \implies \ln|\sec(\pi/2) + \tan(\pi/2)| \to \infty.$

Step 5: Conclusion

The integral diverges to infinity because the upper limit at $x = \pi/2$ causes the term $\ln|\sec x + \tan x|$ to blow up.

Thus, the integral does not converge.

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Would you like further clarification on this or an alternative approach? Here are 5 related questions to expand on this topic:

1. What are the convergence criteria for improper integrals?
2. How does the absolute value function affect integrals?
3. Can we compute a modified version of the integral by restricting the domain?
4. What is the integral of $\sec x$ over a finite range that avoids divergence?
5. How does the behavior of trigonometric functions influence the solvability of integrals?

Tip: When working with piecewise functions or absolute values, carefully analyze the regions where the expressions change to ensure proper integration setup.