A) \[\pi \log 2\]
B) \[\frac{\pi }{\log 2}\]
C) \[\pi \]
D) None of these
Correct Answer: A
Solution :
Let \[I=\int_{0}^{\pi /2}{{{\left( \frac{\theta }{\sin \theta } \right)}^{2}}d\theta }=[-{{\theta }^{2}}\cot \theta ]_{0}^{\pi /2}+\int_{0}^{\pi /2}{\,\,\,\,2\theta .\cot \theta .\,d\theta }\] \[=2[\theta .\log \sin \theta ]_{0}^{\pi /2}-2\int_{0}^{\pi /2}{\log \sin \theta \,d\theta }\] \[\Rightarrow \frac{I}{2}=0-\underset{\theta \to 0}{\mathop{\lim }}\,\theta \log .\sin \theta \]\[-\int_{0}^{\pi /2}{\log \sin \theta \,d\theta }\] Þ \[\frac{\pi }{2}\log 2\]. Hence I =\[\pi \log 2\].
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