Flag Integral Calculus> ∫sinxcosxcos2xcos4xcos8xcos16x dx=?...
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∫sinxcosxcos2xcos4xcos8xcos16x dx=?

UTKARSH CHOUHAN , 10 Years ago
Grade 12
anser 1 Answers
Vikas TU
∫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
 
 
Last Activity: 9 Years ago
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