Question

Evaluate the definite integral: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin |x| + \cos |x|)_~dx \]

Hint:

An alternative way to think about this is that the graph of \(\sin|x| + \cos|x|\) is identical on both the left and right sides of the y-axis. You can just calculate the area for the right side (\(x > 0\)) and multiply it by 2.

Answer 1

Concept: This question requires checking the symmetry properties of an integrand over a symmetric interval \([-a, a]\). The standard definite integral property states: \[ \int_{-a}^{a} f(x) \, dx = \begin{cases} 2 \int_{0}^{a} f(x) \, dx & \text{if } f(x) \text{ is an even function, i.e., } f(-x) = f(x) 0 & \text{if } f(x) \text{ is an odd function, i.e., } f(-x) = -f(x) \end{cases} \] An absolute value function naturally satisfies \(|-x| = |x|\), meaning any function composed strictly of absolute variables behaves symmetrically across the y-axis. Furthermore, within the restricted positive interval of integration \([0, a]\), the expression simplifies because \(|x| = x\).

Step 1: Test whether the integrand behaves as an even or odd function.
Let our complete integrand function be: \[ f(x) = \sin |x| + \cos |x| \] To evaluate its parity, substitute \(-x\) in place of \(x\): \[ f(-x) = \sin |-x| + \cos |-x| \] Since the absolute value removes any negative sign, we know that \(|-x| = |x|\). Substituting this property back: \[ f(-x) = \sin |x| + \cos |x| \] Comparing this result to our original definition, we see that: \[ f(-x) = f(x) \] Therefore, \(f(x)\) is an even function.

Step 2: Apply the even function property to adjust the limits of integration.
Using the property for even functions, we can rewrite our integral as: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) \, dx = 2 \int_{0}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) \, dx \]

Step 3: Simplify the absolute value expression within the positive domain.
For the newly established domain of integration, \(x\) ranges from \(0\) to \(\frac{\pi}{2}\). Since all numbers within \([0, \frac{\pi}{2}]\) are non-negative, the absolute value notation simplifies to: \[ |x| = x \] Substituting this simplification directly into our integral expression: \[ I = 2 \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx \]

Step 4: Execute the integration step and apply boundaries.
We evaluate the anti-derivatives of the fundamental trigonometric components: \[ \int \sin x \, dx = -\cos x \quad \text{and} \quad \int \cos x \, dx = \sin x \] Applying these to our expression: \[ I = 2 \Big[ -\cos x + \sin x \Big]_{0}^{\frac{\pi}{2}} \] Now, substitute the upper boundary \(\frac{\pi}{2}\) and lower boundary \(0\): \[ I = 2 \left( \left( -\cos\frac{\pi}{2} + \sin\frac{\pi}{2} \right) - \left( -\cos 0 + \sin 0 \right) \right) \] We use the known standard trigonometric constant values: \[ \cos\frac{\pi}{2} = 0, \quad \sin\frac{\pi}{2} = 1, \quad \cos 0 = 1, \quad \sin 0 = 0 \] Substituting these precise numerical values back into our equation: \[ I = 2 \Big( (0 + 1) - (-1 + 0) \Big) \] \[ I = 2 \Big( 1 - (-1) \Big) = 2 \Big( 1 + 1 \Big) = 2(2) = 4 \] Thus, the correct numerical output is 4, which matches option (B).