A) \[\frac{1}{5}{{\log }_{e}}|\sin 5x|-\frac{1}{3}{{\log }_{e}}|\sin 3x|+c\]
B) \[\frac{1}{3}{{\log }_{e}}|\sin 3x|-\frac{1}{5}{{\log }_{e}}|\sin 5x|+c\]
C) \[\frac{1}{3}{{\log }_{e}}|\sin 3x|+\frac{1}{5}{{\log }_{e}}|\sin 5x|+c\]
D) \[-\frac{1}{2}\cos 2x+\frac{1}{3}{{\log }_{e}}|\sin 3x|\] \[+\frac{1}{5}{{\log }_{e}}|sin5x|+c\]
E) \[-\frac{1}{2}\cos 2x-\frac{1}{3}{{\log }_{e}}|\sin 3x|\]\[-\frac{1}{5}{{\log }_{e}}|\sin 5x|+c\]
Correct Answer: B
Solution :
\[\int{\frac{\sin 2x}{\sin 3x\sin 5x}}dx=\int{\frac{\sin (5x-3x)}{\sin 3x\sin 5x}}dx\] \[=\int{\frac{\sin 5x\cos 3x-\cos 5x\sin 3x}{\sin 3x\sin 5x}}dx\] \[=\int{(\cot 3x-\cot 5x)}dx\] \[=\frac{1}{3}\log |\sin 3x|-\frac{1}{5}\log |\sin 5x|+c\]
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