Question

Evaluate : \( \int_{-\pi/6}^{\pi/2} (\sin |x| + \cos |x|) dx \)

Hint:

Always handle modulus functions by splitting the domain where the inner function changes sign.
Recall that \( \cos(-x) = \cos x \) and \( \sin(-x) = -\sin x \).

Answer 1

Concept:
• Definition of absolute value: \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \).
• Definite integral property: \( \int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx \).

Step 1: Split the integral at zero
\[ I = \int_{-\pi/6}^{0} (\sin(-x) + \cos(-x)) dx + \int_{0}^{\pi/2} (\sin x + \cos x) dx \]
\[ I = \int_{-\pi/6}^{0} (-\sin x + \cos x) dx + \int_{0}^{\pi/2} (\sin x + \cos x) dx \]

Step 2: Integrate the two parts
For the first part:
\[ [\cos x + \sin x]_{-\pi/6}^{0} = (1 + 0) - \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) = \frac{3 - \sqrt{3}}{2} \]
For the second part:
\[ [-\cos x + \sin x]_{0}^{\pi/2} = (0 + 1) - (-1 + 0) = 2 \]

Step 3: Sum the results
\[ I = \frac{3 - \sqrt{3}}{2} + 2 = \frac{3 - \sqrt{3} + 4}{2} = \frac{7 - \sqrt{3}}{2} \]