Question

Find the value of the integral \( \int_0^{\frac{\pi}{2}} \sin^2(x) \, dx \).

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{6} \)

Hint:

Use the identity for \( \sin^2(x) \) to simplify the integral and break it into simpler parts for easier evaluation.

Answer 1

The Correct Option is A

We are tasked with evaluating the integral:
\[ I = \int_0^{\frac{\pi}{2}} \sin^2(x) \, dx. \] We use the identity for \( \sin^2(x) \):
\[ \sin^2(x) = \frac{1 - \cos(2x)}{2}. \] Thus, the integral becomes:
\[ I = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 - \cos(2x)) \, dx. \] We can now split the integral:
\[ I = \frac{1}{2} \left[ \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \right]. \] Evaluating each integral:
\[ \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}, \quad \int_0^{\frac{\pi}{2}} \cos(2x) \, dx = 0. \] Therefore, the value of the integral is:
\[ I = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}. \]