Question Number 159528 by HongKing last updated on 18/Nov/21
$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{6}}} {\int}}\frac{\mathrm{sin}\left(\mathrm{x}\right)\centerdot\mathrm{sin}\left(\mathrm{x}\:+\:\frac{\pi}{\mathrm{3}}\right)\centerdot\mathrm{sin}\left(\mathrm{x}\:+\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}{\mathrm{sin}\left(\mathrm{3x}\right)\:+\:\mathrm{cos}\left(\mathrm{3x}\right)}\:\mathrm{dx} \\ $$$$\mathrm{Answer}:\:\:\frac{\pi}{\mathrm{48}} \\ $$
Answered by mnjuly1970 last updated on 18/Nov/21
$$\:\:\:\mathrm{I}{dentity}::\:\:{sin}\left({x}\right).{sin}\left(\frac{\pi}{\mathrm{3}}β{x}\right).{sin}\left(\frac{\pi}{\mathrm{3}}\:+{x}\:\right)=\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{x}\right) \\ $$$$\:\:\:{note}:\:\:{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\:+{x}\right)={sin}\left(\piβ\frac{\mathrm{2}\pi}{\mathrm{3}}β{x}\right)={sin}\left(\frac{\pi}{\mathrm{3}}\:β{x}\right) \\ $$$$\:\:\:\Omega\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{6}}} \frac{{sin}\left(\mathrm{3}{x}\right)}{{sin}\left(\mathrm{3}{x}\right)+{cos}\left(\mathrm{3}{x}\right)}{dx} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{12}}\left\{\int\:_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\:{sin}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}\:{dx}\underset{\int_{{a}} ^{\:{b}} {f}\left({a}+{b}β{x}\right){dx}} {\overset{\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}} {=}}\shortparallel\:\:\:\:\:\frac{\pi}{\mathrm{4}}\right\} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{12}}\:\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{48}}\:\:\:…\checkmark \\ $$
Commented by HongKing last updated on 18/Nov/21
$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser} \\ $$
Commented by mnjuly1970 last updated on 18/Nov/21
$$\:\:\:\:{thanks}\:{alot}\:{sir}.. \\ $$