Question Number 86405 by Power last updated on 28/Mar/20
Commented by MJS last updated on 28/Mar/20
$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{3}{x}\:\mathrm{sin}^{\mathrm{3}} \:\mathrm{2}{x}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{sin}\:\mathrm{12}{x}\:−\mathrm{3sin}\:\mathrm{8}{x}\:−\mathrm{2sin}\:\mathrm{6}{x}\:+\mathrm{3sin}\:\mathrm{4}{x}\:+\mathrm{6sin}\:\mathrm{2}{x}\right) \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{super}\:\mathrm{easy} \\ $$
Commented by jagoll last updated on 28/Mar/20
$$\mathrm{wkwkwkwk} \\ $$
Answered by jagoll last updated on 28/Mar/20
![(1/4)(2sin 3x sin 2x)^2 = (1/4)[ cos x−cos 5x ] ∫ (1/4)(sin 2x cos x − sin 2x cos 5x ) dx it now easy to solve](https://www.tinkutara.com/question/Q86413.png)
$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2sin}\:\mathrm{3x}\:\mathrm{sin}\:\mathrm{2x}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\:\mathrm{cos}\:\mathrm{x}−\mathrm{cos}\:\mathrm{5x}\:\right] \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{x}\:−\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{5x}\:\right)\:\mathrm{dx} \\ $$$$\mathrm{it}\:\mathrm{now}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$