A) \[\frac{13}{256}\]
B) \[\frac{15}{64}\]
C) \[\frac{13}{32}\]
D) \[\frac{15}{128}\]
Correct Answer: D
Solution :
\[\int\limits_{\frac{\pi }{12}}^{\frac{\pi }{4}}{\frac{\cos 2x}{{{\left( \frac{1}{\sin 2x} \right)}^{3}}}}=\int\limits_{\frac{\pi }{12}}^{\frac{\pi }{4}}{\cos 2x\times \sin 2x.{{\sin }^{2}}(2x)dx}\] \[=\frac{1}{4}\int_{{}}^{{}}{\sin 4x.(1-cos4x)dx}\] \[=\frac{1}{4}\int\limits_{\frac{\pi }{12}}^{\frac{\pi }{4}}{\sin 4x}-\int\limits_{\frac{\pi }{12}}^{\frac{\pi }{4}}{\sin 8x}\]
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