Question Number 153760 by ZiYangLee last updated on 10/Sep/21
$$\int\:\mathrm{sin}^{\mathrm{2}} \mathrm{4}{x}\:\mathrm{cos}\:\mathrm{4}{x}\:{dx}= \\ $$
Answered by puissant last updated on 10/Sep/21
![=∫sin^2 4x cos4x dx =(1/2)∫sin4x sin8x dx =−(1/4)∫−2sin8x sin4x dx =−(1/4)∫(cos12x+cos4x) dx =−(1/4)[(1/(12))sin12x+(1/4)sin4x]+C =−(1/(48))sin12x−(1/(16))sin4x+C ∴∵ I= −(1/(48))(sin12x+3sin4x)+C..](Q153761.png)
$$=\int{sin}^{\mathrm{2}} \mathrm{4}{x}\:{cos}\mathrm{4}{x}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{sin}\mathrm{4}{x}\:{sin}\mathrm{8}{x}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int−\mathrm{2}{sin}\mathrm{8}{x}\:{sin}\mathrm{4}{x}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int\left({cos}\mathrm{12}{x}+{cos}\mathrm{4}{x}\right)\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{12}}{sin}\mathrm{12}{x}+\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{4}{x}\right]+{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{48}}{sin}\mathrm{12}{x}−\frac{\mathrm{1}}{\mathrm{16}}{sin}\mathrm{4}{x}+{C} \\ $$$$ \\ $$$$\therefore\because\:\:{I}=\:−\frac{\mathrm{1}}{\mathrm{48}}\left({sin}\mathrm{12}{x}+\mathrm{3}{sin}\mathrm{4}{x}\right)+{C}.. \\ $$
Commented by peter frank last updated on 10/Sep/21
$$\mathrm{thanks} \\ $$
Commented by puissant last updated on 10/Sep/21
$${you}'{re}\:{welcome} \\ $$