A) 0
B) \[\pi \ell n2\]
C) \[-\pi \ell n2\]
D) \[\frac{\pi \ell n2}{2}\]
Correct Answer: A
Solution :
[a] Let \[I=\int\limits_{0}^{\frac{\pi }{2}}{\sin 2x\ell n(\cot \,\,x)dx}\] \[=\int\limits_{0}^{\frac{\pi }{2}}{\sin 2x\ell n(\cos \,\,x)dx-\int\limits_{0}^{\frac{\pi }{2}}{\sin 2x\ell n(\sin \,x)dx}}\] \[\int\limits_{0}^{\frac{\pi }{2}}{\sin \left[ 2\left( \frac{\pi }{2}+x \right) \right]\ell n\,\cos \left( \frac{\pi }{2}+x \right)dx}\] \[-\int\limits_{0}^{\frac{\pi }{2}}{\sin 2x\,\ell n(\sin \,\,x)dx}\] \[=\int\limits_{0}^{\frac{\pi }{2}}{\sin (\pi +2x)\ell n(\sin \,x)dx}\] \[-\int\limits_{0}^{\frac{\pi }{2}}{\sin 2x\ell n(\sin \,x)dx}\] \[=\int\limits_{0}^{\frac{\pi }{2}}{\sin 2x\ell n}(\sin x)dx-\int\limits_{0}^{\frac{\pi }{2}}{\sin \,\,2x\,\,\ell n(\sin \,\,x)dx}\] \[=0\]
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