Question Number 102099 by Study last updated on 06/Jul/20

∫sinx ∙ cosx ∙cos2x ∙ cos4x dx=?

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.

Commented byStudy last updated on 06/Jul/20

 (1/8)∫sin8xdx=−(1/(64))cos8x+C right?