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

∫sinx ∙ cosx ∙cos2x ∙ cos4x dx=?

$$\int{sinx}\:\centerdot\:{cosx}\:\centerdot{cos}\mathrm{2}{x}\:\centerdot\:{cos}\mathrm{4}{x}\:{dx}=? \\ $$

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.

$$\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:\mathrm{2xcos}\:\mathrm{2xcos}\:\mathrm{4x}\:\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{sin}\:\mathrm{4xcos}\:\mathrm{4xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{sin}\:\:\mathrm{8xdx}=−\frac{\mathrm{cos}\:\:\mathrm{8x}}{\mathrm{64}}\:+\mathrm{C} \\ $$$$\mathrm{yes}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}. \\ $$

Commented by Study last updated on 06/Jul/20

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

$$\:\frac{\mathrm{1}}{\mathrm{8}}\int{sin}\mathrm{8}{xdx}=−\frac{\mathrm{1}}{\mathrm{64}}{cos}\mathrm{8}{x}+{C}\:{right}? \\ $$

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