MCQ
$\int_0^{\pi /2} {{{\left( {\frac{\theta }{{\sin \theta }}} \right)}^2}d\theta = } $
- ✓$\pi \log 2$
- B$\frac{\pi }{{\log 2}}$
- C$\pi $
- DNone of these
$= [ - {\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 - \mathop {\lim }\limits_{\theta \to 0} \theta \log .\sin \theta $
$ - \int_0^{\pi /2} {\log \sin \theta \,d\theta } $
==> $\frac{\pi }{2}\log 2$.
Hence $I =\pi \log 2$.
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