Question:medium

\(\int_{-\pi/2}^{\pi/2} |\sin x| dx\) is

Show Hint

For even symmetry, double the integral from 0 to limit.
Updated On: Apr 18, 2026
  • 2
  • 0
  • \(\pi/2\)
  • 1
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The absolute value function \( |\sin x| \) is always non-negative. We need to split the integral where the sign of \( \sin x \) changes, or use the property of even functions.
Step 2: Detailed Explanation:
The function \( f(x) = |\sin x| \) is an even function because \( f(-x) = |\sin(-x)| = |-\sin x| = |\sin x| = f(x) \).
For even functions: \[ \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \] So, \( \int_{-\pi/2}^{\pi/2} |\sin x| dx = 2 \int_{0}^{\pi/2} |\sin x| dx \).
In the interval \( [0, \pi/2] \), \( \sin x \) is non-negative, so \( |\sin x| = \sin x \). \[ = 2 \int_{0}^{\pi/2} \sin x dx \] \[ = 2 [-\cos x]_{0}^{\pi/2} \] \[ = 2 [-\cos(\pi/2) - (-\cos(0))] \] \[ = 2 [0 + 1] = 2 \].
Step 3: Final Answer:
The value of the integral is 2.
Was this answer helpful?
0