$$ \int_0^{\frac{\pi}{2}}\!\,\,\sin^{2\text{ }}x\,\mathrm{d}x $$

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.

Learn how to evaluate the integral of sin squared x from 0 to pi/2 using trigonometric identities and definite integrals. This problem covers integration techniques and trigonometric formulas. Suitable for advanced high school students.

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