Question

$ \int{\left( \frac{\sin 2x}{\sin 3x\sin 5x} \right)}dx $ is equal to:

• 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 $

Answer 1

Answer: (B)

$ \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 $