Question

Find the value of \( \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx \).

Hint:

Recognizing whether a function is even or odd is a powerful shortcut for definite integrals over symmetric intervals like \( [-a, a] \). For an even function, the integral is twice the integral over \( [0, a] \). For an odd function, the integral is zero.

Answer 1

Step 1: Understanding the Concept:
We need to evaluate a definite integral of \( \sin^2 x \) over a symmetric interval \( [-\pi/2, \pi/2] \). We can use properties of even and odd functions to simplify the integral. Then, we use a trigonometric identity to integrate \( \sin^2 x \).
Step 2: Key Formula or Approach:
1. Property of definite integrals over symmetric intervals: If \( f(x) \) is an even function (\( f(-x) = f(x) \)), then \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).
2. Power-reducing identity: \( \sin^2 x = \frac{1 - \cos(2x)}{2} \).
Step 3: Detailed Explanation or Calculation:
Let \( f(x) = \sin^2 x \). Let's check if it is an even function: \[ f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x) \] Since \( f(x) \) is an even function, we can simplify the integral: \[ \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = 2 \int_{0}^{\pi/2} \sin^2 x \, dx \] Now, use the power-reducing identity: \[ = 2 \int_{0}^{\pi/2} \frac{1 - \cos(2x)}{2} \, dx = \int_{0}^{\pi/2} (1 - \cos(2x)) \, dx \] Integrate term by term: \[ = \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\pi/2} \] Evaluate at the limits: \[ = \left( \frac{\pi}{2} - \frac{\sin(2 \cdot \pi/2)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \] \[ = \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right) - (0 - 0) \] Since \( \sin(\pi) = 0 \): \[ = \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Step 4: Final Answer:
The value of the integral is \( \frac{\pi}{2} \).