The integral provided is

$\int_{\pi/6}^{\pi/2} |\cos x| \, dx$

To evaluate it without the absolute value, we analyze the behavior of $\cos x$ over the interval $[\pi/6, \pi/2]$. In this range, $\cos x \geq 0$, so the absolute value of $\cos x$ simplifies to $\cos x$ itself.

Now, the integral becomes:

$\int_{\pi/6}^{\pi/2} \cos x \, dx$

The antiderivative of $\cos x$ is $\sin x$, so:

$\int_{\pi/6}^{\pi/2} \cos x \, dx = \sin x \bigg|_{\pi/6}^{\pi/2}$

Substituting the bounds:

$\sin(\pi/2) - \sin(\pi/6)$

We calculate:

$\sin(\pi/2) = 1 \quad \text{and} \quad \sin(\pi/6) = \frac{1}{2}$

Thus:

$1 - \frac{1}{2} = \frac{1}{2}$

The value of the integral is:

$\boxed{\frac{1}{2}}$

Would you like me to explain the steps further or provide additional examples?

Related Questions:

1. How do absolute values affect the integral of a function in general?
2. What if the interval had included a region where $\cos x < 0$? How would that change the solution?
3. Can we generalize this result to any interval where the sign of $\cos x$ is constant?
4. How does the Fundamental Theorem of Calculus relate to solving this problem?
5. Could this integral be solved numerically for more complex cases?

Tip:

Always check the behavior of the function under the absolute value to determine whether splitting the integral is necessary.