We want to evaluate the integral of $3 \cos^2(x)$ from $0$ to $\frac{\pi}{2}$. The integral can be written as:

$\int_0^{\frac{\pi}{2}} 3 \cos^2(x) \, dx$

We will first use a trigonometric identity to simplify $\cos^2(x)$. The identity is:

$\cos^2(x) = \frac{1 + \cos(2x)}{2}$

Substituting this into the integral, we get:

$\int_0^{\frac{\pi}{2}} 3 \cdot \frac{1 + \cos(2x)}{2} \, dx$

Simplifying:

$\frac{3}{2} \int_0^{\frac{\pi}{2}} (1 + \cos(2x)) \, dx$

Now, split this into two separate integrals:

$\frac{3}{2} \left[ \int_0^{\frac{\pi}{2}} 1 \, dx + \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \right]$

First integral:

$\int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}$

Second integral:

$\int_0^{\frac{\pi}{2}} \cos(2x) \, dx$

To solve this, use substitution: Let $u = 2x$, so $du = 2 dx$, or $dx = \frac{du}{2}$.

The limits change as well: when $x = 0$, $u = 0$; when $x = \frac{\pi}{2}$, $u = \pi$.

So the integral becomes:

$\frac{1}{2} \int_0^{\pi} \cos(u) \, du$

We know that $\int \cos(u) \, du = \sin(u)$, so:

[ \frac{1}{2} \left[ \sin(u) \right]_0^{\pi} = \frac{1}{2} \left[ \sin(\pi) - \sin(0) \right] = \frac{1}{2} \cdot (0 - 0) = 0 ]

So, the second integral evaluates to 0.

Final result:

Now, combining the two results:

$\frac{3}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{3}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{4}$

Thus, the value of the integral is:

$\boxed{\frac{3\pi}{4}}$

Would you like further details or have any questions?

Here are 5 related questions:

1. How would the integral change if the bounds were different?
2. What other trigonometric identities can simplify similar integrals?
3. How would you integrate $\cos^4(x)$ over the same bounds?
4. How can substitution help with more complex integrals?
5. What is the significance of definite integrals in calculating areas?

Tip: For trigonometric integrals, always check if you can use identities to simplify powers of sine or cosine.