Question Number 131068 by EDWIN88 last updated on 01/Feb/21
$$\:\int\:\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:\mathrm{4}{x}\:{dx}\:=?\: \\ $$
Answered by Ar Brandon last updated on 01/Feb/21
![I=∫sin^2 xcos4xdx=(1/2)∫(1+cos2x)cos4xdx =(1/2)∫[cos4x+(1/2)(cos6x+cos2x)]dx =((sin4x)/8)+((sin6x)/(24))+((sin2x)/8)+C](https://www.tinkutara.com/question/Q131069.png)
$$\mathcal{I}=\int\mathrm{sin}^{\mathrm{2}} \mathrm{xcos4xdx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos2x}\right)\mathrm{cos4xdx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\mathrm{cos4x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos6x}+\mathrm{cos2x}\right)\right]\mathrm{dx} \\ $$$$\:\:\:=\frac{\mathrm{sin4x}}{\mathrm{8}}+\frac{\mathrm{sin6x}}{\mathrm{24}}+\frac{\mathrm{sin2x}}{\mathrm{8}}+\mathcal{C} \\ $$