∫sinxcosxcos2xcos4xcos8xcos16x dx=?
UTKARSH CHOUHAN , 9 Years ago
Grade 12
Integral Calculus> ∫sinxcosxcos2xcos4xcos8xcos16x dx=?...
1 AnswersVikas TU
Last Activity: 8 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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