Question Number 75924 by benjo last updated on 21/Dec/19
$$\mathrm{Thr}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\underset{\mathrm{0}\:} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{6}}\right){dx}\:\:\mathrm{is} \\ $$
Commented by mathmax by abdo last updated on 21/Dec/19
$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{{sin}\left({x}−\frac{\pi}{\mathrm{3}}\right){sin}\left({x}−\frac{\pi}{\mathrm{6}}\right)}\:\:\:{but}\:{we}\:{have} \\ $$$${sin}\left({x}−\frac{\pi}{\mathrm{3}}\right){sin}\left({x}−\frac{\pi}{\mathrm{6}}\right)\:={cos}\left(\frac{\pi}{\mathrm{2}}−{x}+\frac{\pi}{\mathrm{3}}\right){cos}\left(\frac{\pi}{\mathrm{2}}−{x}\:+\frac{\pi}{\mathrm{6}}\right) \\ $$$$={cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−{x}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}+\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{2}{x}\right)\:+{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−{x}−\frac{\mathrm{2}\pi}{\mathrm{3}}+{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{2}{x}\right)\:+{cos}\left(\frac{\pi}{\mathrm{6}}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(−\frac{\pi}{\mathrm{2}}−\mathrm{2}{x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{sin}\left(\mathrm{2}{x}\right)\right\}\:\Rightarrow{I}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)}\:=_{\mathrm{2}{x}={t}} \mathrm{4}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{2}{sint}\right)} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{\:\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left({t}\right)}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{2}\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\mathrm{4}{u}}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\:\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\mathrm{4}{u}\:+\sqrt{\mathrm{3}}} \\ $$$$\Delta^{'} \:=\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\:=\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =\frac{\mathrm{2}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}\:=\sqrt{\mathrm{3}} \\ $$$${u}_{\mathrm{2}} =\frac{\mathrm{2}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow{I}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\:\sqrt{\mathrm{3}}\left({u}−\sqrt{\mathrm{3}}\right)\left({u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\int_{\mathrm{0}} ^{\infty} \:\:\left(\frac{\mathrm{1}}{{u}−\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{{u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right){du} \\ $$$$=\mathrm{2}\:\:{ln}\mid\frac{{u}−\sqrt{\mathrm{3}}}{{u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\mid\:+{c} \\ $$$$=\mathrm{2}{ln}\mid\frac{{tan}\left({x}\right)−\sqrt{\mathrm{3}}}{{tan}\left({x}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\mid\:+{c} \\ $$
Commented by benjo last updated on 22/Dec/19
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Commented by abdomathmax last updated on 23/Dec/19
$${you}\:{are}\:{welcome}. \\ $$