Question

Evaluate \[ \int_{\pi/4}^{5\pi/4}\left(|\cos t|\sin t+|\sin t|\cos t\right)\,dt \]

• A. \(0\)
• B. \(1\)
• C. \(\dfrac{1}{2}\)
• D. \(\dfrac{\sqrt{3}}{2}\)

Hint:

For integrals involving modulus functions, always split the interval wherever the sign of the trigonometric function changes.

Answer 1

The Correct Option is A

Step 1: Split the interval according to signs of \(\sin t\) and \(\cos t\).
The interval is \[ \left[\frac{\pi}{4},\frac{5\pi}{4}\right] \] We split it into two parts: \[ \left[\frac{\pi}{4},\frac{\pi}{2}\right] \quad \text{and} \quad \left[\frac{\pi}{2},\frac{5\pi}{4}\right] \] In the interval \[ \left[\frac{\pi}{4},\frac{\pi}{2}\right], \] both \(\sin t\) and \(\cos t\) are positive. Hence, \[ |\sin t|=\sin t,\qquad |\cos t|=\cos t \] Therefore, the integrand becomes \[ \cos t\sin t+\sin t\cos t \] \[ =2\sin t\cos t \] \[ =\sin 2t \]

Step 2: Evaluate the first integral.
\[ I_1=\int_{\pi/4}^{\pi/2}\sin 2t\,dt \] \[ =\left[-\frac{\cos 2t}{2}\right]_{\pi/4}^{\pi/2} \] \[ = -\frac{\cos \pi}{2} + \frac{\cos \frac{\pi}{2}}{2} \] \[ = -\frac{-1}{2}+0 \] \[ =\frac12 \]

Step 3: Evaluate the second interval.
For \[ t\in\left[\frac{\pi}{2},\pi\right], \] we have \[ \sin t\gt 0,\qquad \cos t\lt 0 \] Thus, \[ |\sin t|=\sin t,\qquad |\cos t|=-\cos t \] Hence, \[ |\cos t|\sin t+|\sin t|\cos t \] \[ = (-\cos t)\sin t+(\sin t)\cos t \] \[ =0 \] Similarly, for \[ t\in\left[\pi,\frac{5\pi}{4}\right], \] both \(\sin t\) and \(\cos t\) are negative. Hence, \[ |\sin t|=-\sin t,\qquad |\cos t|=-\cos t \] Therefore, \[ (-\cos t)\sin t+(-\sin t)\cos t \] \[ =-2\sin t\cos t \] \[ =-\sin 2t \] Thus, \[ I_2=\int_{\pi}^{5\pi/4} -\sin 2t\,dt \] \[ = \left[\frac{\cos 2t}{2}\right]_{\pi}^{5\pi/4} \] \[ = \frac{\cos \frac{5\pi}{2}}{2} - \frac{\cos 2\pi}{2} \] \[ = 0-\frac12 \] \[ =-\frac12 \]

Step 4: Add all parts.
Total integral is \[ I=I_1+0+I_2 \] \[ =\frac12-\frac12 \] \[ =0 \]

Step 5: Final conclusion.
Hence, \[ \boxed{0} \]