In the image, I can see the following integral problem written on the board:

$\int_0^{\frac{\pi}{2}} \sqrt{\sin{x}} \cdot \cos{x} \, dx$

Let's solve this step by step.

Step 1: Substitution

We will use the substitution method to simplify the integral. Set: $u = \sin{x}$ Thus, $du = \cos{x} \, dx$.

Step 2: Adjust Limits of Integration

When $x = 0$, $u = \sin{0} = 0$. When $x = \frac{\pi}{2}$, $u = \sin{\frac{\pi}{2}} = 1$.

Step 3: Rewrite the Integral

Substitute $u$ into the integral: $\int_0^{1} \sqrt{u} \, du$

Step 4: Evaluate the Integral

The integral of $\sqrt{u}$ is: $\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$

Step 5: Apply Limits of Integration

Now, evaluate this from 0 to 1: $\left[ \frac{2}{3} u^{3/2} \right]_0^1 = \frac{2}{3} (1^{3/2} - 0^{3/2}) = \frac{2}{3}$

Thus, the value of the integral is $\frac{2}{3}$.

Would you like more details on any step or have any questions?

Here are 5 related questions to further explore this topic:

1. How do you choose appropriate substitutions for integrals?
2. What is the process of changing the limits of integration when making a substitution?
3. How do you evaluate more complex integrals with multiple substitutions?
4. What are common trigonometric substitutions for integrals involving sine and cosine?
5. How does the integration of powers of trigonometric functions differ from basic polynomial integration?

Tip: Always double-check the limits of integration after substitution to ensure the bounds align with the new variable!