Find:   
      \int \left ( \sin x\cdot \sin 2x\cdot \sin 3x \right )dx

 

 

 

 
 
 
 
 

Answers (1)

\int \left ( \sin x\cdot \sin 2x\cdot \sin 3x \right )dx= \frac{1}{2}\int \left ( 2\sin x\cdot \sin 2x \right )\sin 3x\, dx
       = \frac{1}{2}\int \left [ \cos x-\cos 3x \right ]\sin 3x\, dx
      = \frac{1}{4}\int \left [ 2\sin 3x\, \cos x-2\cos 3x\sin 3x \right ]dx
      = \frac{1}{4}\int \left [\sin 4x+\sin 2x-\sin 6x \right ]dx
      = \frac{1}{4} \left [\frac{1}{6}\cos 6x -\frac{1}{4}\cos 4x-\frac{1}{2}\cos 2x\right ]+c
      = \frac{1}{24} \left [\cos 6x\right ]\frac{-1}{16}\cos 4x-\frac{1}{8}\cos 2x+c
      
       

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