A) \[\log \sin 3x-\log \sin 5x+c\]
B) \[\frac{1}{3}\log \sin 3x+\frac{1}{5}\log \sin 5x+c\]
C) \[\frac{1}{3}\log \sin 3x-\frac{1}{5}\log \sin 5x+c\]
D) \[3\log \sin 3x-5\log \sin 5x+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{\sin 2x}{\sin 5x\sin 3x}\,dx}=\int_{{}}^{{}}{\frac{\sin (5x-3x)}{\sin 5x\sin 3x}\,dx}\] \[=\int_{{}}^{{}}{\frac{\sin 5x\cos 3x-\cos 5x\sin 3x}{\sin 5x\sin 3x}\,dx}\] \[=\frac{1}{3}\log \sin 3x-\frac{1}{5}\log \sin 5x+c.\]
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