If int_3^5 sqrt{8 x-x^2-15} d x=p,then sin p+ | vedclass

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(D) Let $I = \int_3^5 \sqrt{8x - x^2 - 15} dx$.We can rewrite the quadratic expression as $-(x^2 - 8x + 15) = -(x^2 - 8x + 16 - 1) = 1 - (x - 4)^2$.Th

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$2$

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(D) Let $I = \int_3^5 \sqrt{8x - x^2 - 15} dx$.We can rewrite the quadratic expression as $-(x^2 - 8x + 15) = -(x^2 - 8x + 16 - 1) = 1 - (x - 4)^2$.Thus,$I = \int_3^5 \sqrt{1 - (x - 4)^2} dx$.Let $x - 4 = \sin 	heta$,then $dx = \cos 	heta d	heta$.When $x = 3$,$\sin 	heta = -1$,so $	heta = -\frac{\pi}{2}$.When $x = 5$,$\sin 	heta = 1$,so $	heta = \frac{\pi}{2}$.$I = \int_{-\pi/2}^{\pi/2} \sqrt{1 - \sin^2 	heta} \cos 	heta d	heta = \int_{-\pi/2}^{\pi/2} \cos^2 	heta d	heta$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get $I = \int_{-\pi/2}^{\pi/2} \frac{1 + \cos 2	heta}{2} d	heta = \left[ \frac{	heta}{2} + \frac{\sin 2	heta}{4} ight]_{-\pi/2}^{\pi/2} = (\frac{\pi}{4} + 0) - (-\frac{\pi}{4} + 0) = \frac{\pi}{2}$.So,$p = \frac{\pi}{2}$.Then $\sin p + \operatorname{cosec} p = \sin(\frac{\pi}{2}) + \operatorname{cosec}(\frac{\pi}{2}) = 1 + 1 = 2$.

If $\int_3^5 \sqrt{8 x-x^2-15} d x=p$,then $\sin p+\operatorname{cosec} p=$

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