intlimits_{frac{pi }{6}}^{frac{5pi }{6}} {lef | vedclass

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Question

(B) ધારો કે $I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2} [9\sin^2 	heta - (1 + 2\sin 	heta + \sin^2 	heta)] \, d	heta$ $I = \frac{1}{2}

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(B) ધારો કે $I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2} [9\sin^2 	heta - (1 + 2\sin 	heta + \sin^2 	heta)] \, d	heta$ $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (8\sin^2 	heta - 2\sin 	heta - 1) \, d	heta$ $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ નો ઉપયોગ કરતા: $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [8(\frac{1 - \cos 2	heta}{2}) - 2\sin 	heta - 1] \, d	heta$ $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (4 - 4\cos 2	heta - 2\sin 	heta - 1) \, d	heta$ $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (3 - 4\cos 2	heta - 2\sin 	heta) \, d	heta$ $I = \frac{1}{2} [3	heta - 2\sin 2	heta + 2\cos 	heta]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}$ $I = \frac{1}{2} [ (3(\frac{5\pi}{6}) - 2\sin(\frac{5\pi}{3}) + 2\cos(\frac{5\pi}{6})) - (3(\frac{\pi}{6}) - 2\sin(\frac{\pi}{3}) + 2\cos(\frac{\pi}{6})) ]$ $I = \frac{1}{2} [ (\frac{5\pi}{2} - 2(-\frac{\sqrt{3}}{2}) + 2(-\frac{\sqrt{3}}{2})) - (\frac{\pi}{2} - 2(\frac{\sqrt{3}}{2}) + 2(\frac{\sqrt{3}}{2})) ]$ $I = \frac{1}{2} [ (\frac{5\pi}{2} + \sqrt{3} - \sqrt{3}) - (\frac{\pi}{2} - \sqrt{3} + \sqrt{3}) ]$ $I = \frac{1}{2} [ \frac{5\pi}{2} - \frac{\pi}{2} ] = \frac{1}{2} [ 2\pi ] = \pi$.

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

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