To evaluate the integral $\int_0^{\frac{\pi}{2}} \sin^2 x \, \mathrm{d}x$, we can use a trigonometric identity to simplify the integrand.

Recall the identity: $\sin^2 x = \frac{1 - \cos(2x)}{2}$

Thus, the integral becomes: $\int_0^{\frac{\pi}{2}} \sin^2 x \, \mathrm{d}x = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, \mathrm{d}x$

We can split this into two separate integrals: $\int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, \mathrm{d}x = \frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, \mathrm{d}x - \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2x) \, \mathrm{d}x$

Now, evaluate each integral separately.

1. The first integral: $\frac{1}{2} \int_0^{\frac{\pi}{2}} 1 \, \mathrm{d}x = \frac{1}{2} \left[ x \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}$

2. The second integral: $\frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2x) \, \mathrm{d}x$

Let $u = 2x$, hence $du = 2 \, dx$ and $dx = \frac{du}{2}$. Adjust the limits of integration accordingly: When $x = 0$, $u = 0$. When $x = \frac{\pi}{2}$, $u = \pi$.

So, the integral becomes: $\frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2x) \, \mathrm{d}x = \frac{1}{2} \cdot \frac{1}{2} \int_0^{\pi} \cos(u) \, \mathrm{d}u = \frac{1}{4} \left[ \sin(u) \right]_0^{\pi}$

$\frac{1}{4} \left( \sin(\pi) - \sin(0) \right) = \frac{1}{4} \left( 0 - 0 \right) = 0$

Therefore: $\int_0^{\frac{\pi}{2}} \sin^2 x \, \mathrm{d}x = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

So, the final answer is: $\boxed{\frac{\pi}{4}}$

Would you like more details on any step or have any other questions? Here are 5 questions you might be interested in:

1. How do you derive the trigonometric identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$?
2. Can you explain more about changing the limits of integration?
3. How do you evaluate definite integrals?
4. What are other methods to solve $\int \sin^2 x \, \mathrm{d}x$?
5. How is the substitution method used in integrals?

Tip: When dealing with trigonometric integrals, using identities to simplify the integrand can make the integral much easier to evaluate.