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sinx-cosx-cos2x-cos4x-dx-




Question Number 102099 by Study last updated on 06/Jul/20
∫sinx ∙ cosx ∙cos2x ∙ cos4x dx=?
sinxcosxcos2xcos4xdx=?
Answered by PRITHWISH SEN 2 last updated on 06/Jul/20
(1/2)∫sin 2xcos 2xcos 4x dx=(1/4)∫sin 4xcos 4xdx  =(1/8)∫sin  8xdx=−((cos  8x)/(64)) +C  yes you are right.
12sin2xcos2xcos4xdx=14sin4xcos4xdx=18sin8xdx=cos8x64+Cyesyouareright.
Commented by Study last updated on 06/Jul/20
 (1/8)∫sin8xdx=−(1/(64))cos8x+C right?
18sin8xdx=164cos8x+Cright?

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