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∫sinxcosxcos2xcos4xcos8xcos16x dx=?

∫sinxcosxcos2xcos4xcos8xcos16x dx=?
 

Grade:12

1 Answers

Vikas TU
14149 Points
7 years ago
∫sinxcosxcos2xcos4xcos8xcos16x dx= > 1/2(∫sin2xcos2xcos4xcos8xcos16x )dx
=> (1/4)∫sin4xcos4xcos8xcos16x dx
=>  (1/8)∫sin8xcos8xcos16x dx
=>  (1/16)∫sin16xcos16x dx
=>  (1/32)∫sin32x dx
=>  -(1/32*32)(cos32x) + c
 
 

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