Step 1: Understanding the Question:
The task is to evaluate a definite integral of \(\sin^2 x\) with limits from 0 to \(\frac{\pi}{2}\). This is a standard integral in calculus.
Step 2: Key Formula or Approach:
Direct integration of \(\sin^2 x\) is not simple. We need to use a trigonometric identity to reduce the power of the sine function. The relevant power-reduction formula is derived from the double-angle identity for cosine:
\[
\cos(2x) = 1 - 2\sin^2 x
\]
Rearranging this gives:
\[
\sin^2 x = \frac{1 - \cos(2x)}{2}
\]
We will substitute this into the integral.
Step 3: Detailed Explanation:
Substitute the identity into the integral:
\[
\int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx
\]
We can split the integral into two parts for easier calculation:
\[
= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx
\]
Now, evaluate each integral separately:
For the first part:
\[
\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{1}{2} [x]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left(\frac{\pi}{2} - 0\right) = \frac{\pi}{4}
\]
For the second part:
\[
\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx = \frac{1}{2} \left[\frac{\sin(2x)}{2}\right]_{0}^{\frac{\pi}{2}} = \frac{1}{4} [\sin(2x)]_{0}^{\frac{\pi}{2}}
\]
Applying the limits:
\[
= \frac{1}{4} (\sin(2 \cdot \frac{\pi}{2}) - \sin(2 \cdot 0)) = \frac{1}{4} (\sin(\pi) - \sin(0)) = \frac{1}{4}(0 - 0) = 0
\]
Combine the results of the two parts:
\[
\int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{4} - 0 = \frac{\pi}{4}
\]
Step 4: Final Answer:
The value of the definite integral is \( \dfrac{\pi}{4} \), which corresponds to option (B).