int_0^{pi / 2} frac{2 sin (x)+3 cos (x)}{sin | vedclass

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(C) Let $I = \int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we ge

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$\frac{5 \pi}{4}$

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(C) Let $I = \int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get: $I = \int_0^{\pi / 2} \frac{2 \sin (\pi/2-x)+3 \cos (\pi/2-x)}{\sin (\pi/2-x)+\cos (\pi/2-x)} d x = \int_0^{\pi / 2} \frac{2 \cos (x)+3 \sin (x)}{\cos (x)+\sin (x)} d x$. Adding the two expressions for $I$: $2I = \int_0^{\pi / 2} \frac{(2 \sin x + 3 \cos x) + (2 \cos x + 3 \sin x)}{\sin x + \cos x} d x$. $2I = \int_0^{\pi / 2} \frac{5 \sin x + 5 \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 5 d x$. $2I = [5x]_0^{\pi / 2} = 5(\frac{\pi}{2} - 0) = \frac{5\pi}{2}$. Therefore,$I = \frac{5\pi}{4}$.

$\int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x=$

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