Question

\(\int_{0}^{20\pi} (|\sin x| + |\cos x|)^2 \,dx\)
is equal to

• A. \(10(π+4)\)
• B. \(10(π+2)\)
• C. \(20(π-2)\)
• D. \(20(π+2)\)

Answer 1

The Correct Option is D

To solve the integral \(\int_{0}^{20\pi} (|\sin x| + |\cos x|)^2 \,dx\), we begin by understanding the periodicity and behavior of the function involved.

1. The functions \(|\sin x|\) and \(|\cos x|\) are both periodic with a period of \(\pi\). Therefore, the function \((|\sin x| + |\cos x|)^2\) will also have a period of \(\pi\).
2. To find the integral over \([0, 20\pi]\), it is sufficient to compute it over one period \([0, \pi]\) and multiply the result by the number of periods, which in this case is 20.
3. First, compute the integral over one period: \[ \int_{0}^{\pi} (|\sin x| + |\cos x|)^2 \, dx \] For \(x \in [0, \pi/2]\), \(|\sin x| = \sin x\) and \(|\cos x| = \cos x\). For \(x \in [\pi/2, \pi]\), \(|\sin x| = \sin x\) and \(|\cos x| = -\cos x\).
4. Calculate separately for both cases:
   • From \(0\) to \(\pi/2\), \[ \int_{0}^{\pi/2} (\sin x + \cos x)^2 \, dx \] Expanding and integrating: \[ = \int_{0}^{\pi/2} (\sin^2 x + 2\sin x \cos x + \cos^2 x) \, dx = \int_{0}^{\pi/2} (1 + 2\sin x \cos x) \, dx ] = \frac{\pi}{2} + 0 = \frac{\pi}{2} \]
   • From \(\pi/2\) to \(\pi\), \[ \int_{\pi/2}^{\pi} (\sin x - \cos x)^2 \, dx \] Similarly, expanding: \[ = \int_{\pi/2}^{\pi} (\sin^2 x - 2\sin x \cos x + \cos^2 x) \, dx = \int_{\pi/2}^{\pi} (1 - 2\sin x \cos x) \, dx ] = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]
5. Thus, the integral over \([0, \pi]\) is: \[ \int_{0}^{\pi} (|\sin x| + |\cos x|)^2 \, dx = \frac{\pi}{2} + \frac{\pi}{2} = \pi + 2 \]
6. Finally, multiply by 20 to account for the full range: \[ \int_{0}^{20\pi} (|\sin x| + |\cos x|)^2 \, dx = 20 \times (\pi + 2) = 20(\pi + 2) \]

Hence, the correct answer is \(20(\pi + 2)\).