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

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(A) माना $I = \int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt$. यहाँ $t \in [\frac{\pi}{4}, \frac{\pi}{2}]$ के लिए,$\cos

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(A) माना $I = \int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt$. यहाँ $t \in [\frac{\pi}{4}, \frac{\pi}{2}]$ के लिए,$\cos t > 0$ और $\sin t > 0$,इसलिए $f(t) = \sin(2t)$. $t \in [\frac{\pi}{2}, \pi]$ के लिए,$\cos t  0$,इसलिए $f(t) = 0$. $t \in [\pi, \frac{5\pi}{4}]$ के लिए,$\cos t अतः,$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2t) dt + \int_{\pi}^{\frac{5\pi}{4}} -\sin(2t) dt$. $I = \left[ -\frac{\cos(2t)}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} + \left[ \frac{\cos(2t)}{2} \right]_{\pi}^{\frac{5\pi}{4}}$. $I = (\frac{1}{2} - 0) + (0 - \frac{1}{2}) = 0$.

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

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