int_{frac{pi}{4}}^{frac{5 pi}{4}} (|cos t| si | vedclass

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(A) Let $I = \int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt$. We know that $\frac{d}{dt} (\sin t \cos t) = \cos^2 t - \

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(A) Let $I = \int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt$. We know that $\frac{d}{dt} (\sin t \cos t) = \cos^2 t - \sin^2 t = \cos(2t)$. However,it is easier to use the property $\int f(t) dt = \int f(t) dt$. Note that the integrand is $f(t) = |\cos t| \sin t + |\sin t| \cos t$. For $t \in [\frac{\pi}{4}, \frac{\pi}{2}]$,$\cos t > 0$ and $\sin t > 0$,so $f(t) = \cos t \sin t + \sin t \cos t = 2 \sin t \cos t = \sin(2t)$. For $t \in [\frac{\pi}{2}, \pi]$,$\cos t  0$,so $f(t) = -\cos t \sin t + \sin t \cos t = 0$. For $t \in [\pi, \frac{5\pi}{4}]$,$\cos t Thus,$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2t) dt + \int_{\frac{\pi}{2}}^{\pi} 0 dt + \int_{\pi}^{\frac{5\pi}{4}} -\sin(2t) dt$. $I = \left[ -\frac{\cos(2t)}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} + 0 + \left[ \frac{\cos(2t)}{2} \right]_{\pi}^{\frac{5\pi}{4}}$. $I = (-\frac{\cos(\pi)}{2} - (-\frac{\cos(\frac{\pi}{2})}{2})) + (\frac{\cos(\frac{5\pi}{2})}{2} - \frac{\cos(2\pi)}{2})$. $I = (\frac{1}{2} - 0) + (0 - \frac{1}{2}) = 0$.

$\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt =$

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