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Question Number 75924 by benjo last updated on 21/Dec/19

Thr value of ∫_(0 ) ^(π/2) cosec (x−(π/3))cosec (x−(π/6))dx is

$$\mathrm{Thr}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\underset{\mathrm{0}\:} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{6}}\right){dx}\:\:\mathrm{is} \\ $$

Commented by mathmax by abdo last updated on 21/Dec/19

I =∫_0 ^(π/2) (dx/(sin(x−(π/3))sin(x−(π/6)))) but we have sin(x−(π/3))sin(x−(π/6)) =cos((π/2)−x+(π/3))cos((π/2)−x +(π/6)) =cos(((5π)/6)−x)cos(((2π)/3)−x) =(1/2){ cos(((5π)/6)+((2π)/3)−2x) +cos(((5π)/6)−x−((2π)/3)+x)} =(1/2){cos(((3π)/2)−2x) +cos((π/6))}=(1/2){cos(−(π/2)−2x)+((√3)/2)} =(1/2){((√3)/2)−sin(2x)} ⇒I =2 ∫_0 ^(π/2) (dx/(((√3)/2)−sin(2x))) =4 ∫_0 ^(π/2) (dx/((√3)−2sin(2x))) =_(2x=t) 4 ∫_0 ^π (dt/(2((√3)−2sint))) =2 ∫_0 ^π (dt/((√3)−2sin(t))) =_(tan((t/2))=u) 2∫_0 ^(+∞) (1/((√3)−2((2u)/(1+u^2 ))))((2du)/(1+u^2 )) =4 ∫_0 ^∞ (du/((√3)+(√3)u^2 −4u)) =4 ∫_0 ^∞ (du/((√3)u^2 −4u +(√3))) Δ^′ =(−2)^2 −3 =1 ⇒u_1 =((2+1)/(√3)) =(3/(√3)) =(√3) u_2 =((2−1)/(√3)) =(1/(√3)) ⇒I =4 ∫_0 ^∞ (du/((√3)(u−(√3))(u−(1/(√3))))) =(4/(√3))×(1/((√3)−(1/(√3))))∫_0 ^∞ ((1/(u−(√3)))−(1/(u−(1/(√3)))))du =2 ln∣((u−(√3))/(u−(1/(√3))))∣ +c =2ln∣((tan(x)−(√3))/(tan(x)−(1/(√3))))∣ +c

$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{{sin}\left({x}−\frac{\pi}{\mathrm{3}}\right){sin}\left({x}−\frac{\pi}{\mathrm{6}}\right)}\:\:\:{but}\:{we}\:{have} \\ $$$${sin}\left({x}−\frac{\pi}{\mathrm{3}}\right){sin}\left({x}−\frac{\pi}{\mathrm{6}}\right)\:={cos}\left(\frac{\pi}{\mathrm{2}}−{x}+\frac{\pi}{\mathrm{3}}\right){cos}\left(\frac{\pi}{\mathrm{2}}−{x}\:+\frac{\pi}{\mathrm{6}}\right) \\ $$$$={cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−{x}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}+\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{2}{x}\right)\:+{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}−{x}−\frac{\mathrm{2}\pi}{\mathrm{3}}+{x}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{2}{x}\right)\:+{cos}\left(\frac{\pi}{\mathrm{6}}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(−\frac{\pi}{\mathrm{2}}−\mathrm{2}{x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{sin}\left(\mathrm{2}{x}\right)\right\}\:\Rightarrow{I}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)}\:=_{\mathrm{2}{x}={t}} \mathrm{4}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{2}{sint}\right)} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left({t}\right)}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}−\mathrm{2}\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\mathrm{4}{u}}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\mathrm{4}{u}\:+\sqrt{\mathrm{3}}} \\ $$$$\Delta^{'} \:=\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\:=\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =\frac{\mathrm{2}+\mathrm{1}}{\sqrt{\mathrm{3}}}\:=\frac{\mathrm{3}}{\sqrt{\mathrm{3}}}\:=\sqrt{\mathrm{3}} \\ $$$${u}_{\mathrm{2}} =\frac{\mathrm{2}−\mathrm{1}}{\sqrt{\mathrm{3}}}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\Rightarrow{I}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\sqrt{\mathrm{3}}\left({u}−\sqrt{\mathrm{3}}\right)\left({u}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\int_{\mathrm{0}} ^{\infty} \:\:\left(\frac{\mathrm{1}}{{u}−\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{{u}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\right){du} \\ $$$$=\mathrm{2}\:\:{ln}\mid\frac{{u}−\sqrt{\mathrm{3}}}{{u}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\mid\:+{c} \\ $$$$=\mathrm{2}{ln}\mid\frac{{tan}\left({x}\right)−\sqrt{\mathrm{3}}}{{tan}\left({x}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\mid\:+{c} \\ $$

Commented by benjo last updated on 22/Dec/19

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Commented by abdomathmax last updated on 23/Dec/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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