A) \[\frac{\pi }{3}+\frac{1}{6}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{2\pi }{3}\]
D) \[\frac{\pi }{9}\]
Correct Answer: B
Solution :
[b] \[\because \,\,\,\,2{{\cot }^{2}}\theta -\frac{5}{\sin \,\theta }+4=0\] \[2+2\,\text{cose}{{\text{c}}^{2}}\,\theta -5\,\text{cosec}\,\theta =0\] \[2\,\text{cose}{{\text{c}}^{2}}\,\theta -4\,\text{cosec}\,\theta -\,\text{cosec}\theta +2=0\] \[\therefore \,\,\,(2\,\text{cosec}\,\,\theta -1)\,(\,\text{cosec}\,\theta -2)\,=0\] \[\therefore \,\,\text{cosec}\,\theta =\frac{1}{2}\] or 2. \[\therefore \,\,\sin \,\theta =2\] or \[\frac{1}{2}.\] \[\because \,\,\theta \,\in \,(0,2\pi )\] \[\therefore \,\,\,\,{{\theta }_{1}}=\frac{\pi }{6}\] and \[{{\theta }_{2}}=\frac{5\pi }{6},\] \[\because \,{{\theta }_{1}}<{{\theta }_{2}}\] \[\therefore \,\,I=\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}}{{{\cos }^{2}}\,3\theta \,d\theta =\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}}{\frac{1+\cos \,6\theta }{2}}}\,d\theta \] \[=\frac{1}{2}\left[ \theta +\frac{\sin \,6\theta }{6} \right]_{\frac{\pi }{6}}^{\frac{5\pi }{6}}=\frac{\pi }{3}\]
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