Question Number 63711 by mathmax by abdo last updated on 07/Jul/19
$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\sqrt{\mathrm{3}}{cosx}\:+\sqrt{\mathrm{2}}{sinx}} \\ $$
Commented by mathmax by abdo last updated on 08/Jul/19
![let A =∫_0 ^π (dx/((√3)cosx +(√2)sinx)) changement tan((x/2))=t give A =∫_0 ^∞ ((2dt)/((1+t^2 ){(√3)((1−t^2 )/(1+t^2 )) +(√2)((2t)/(1+t^2 ))})) =∫_0 ^∞ ((2dt)/((√3)(1−t^2 )+2(√2)t)) =∫_0 ^∞ ((2dt)/(−(√3)t^2 +2(√2)t +(√3))) =∫_0 ^∞ ((−2dt)/((√3)t^2 −2(√2)t−(√3))) Δ^′ =(−(√2))^2 +3 =5 ⇒t_1 =(((√2)+(√5))/(√3)) and t_2 =(((√2)−(√5))/(√3)) ⇒ A =((−2)/(√3))∫_0 ^∞ (dt/((t−t_1 )(t−t_2 ))) =((−2)/((√3)(t_1 −t_2 ))) ∫_0 ^∞ {(1/(t−t_1 )) −(1/(t−t_2 ))}dt =((−2)/((√3)((2(√5))/(√3)))) ∫_0 ^∞ {(1/(t−t_1 )) −(1/(t−t_2 ))}dt=−(1/(√5))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^(+∞) =(1/(√5))ln∣(t_1 /t_2 )∣ =(1/(√5))ln∣(((√2)+(√5))/((√2)−(√5)))∣ =(1/(√5))ln((((√5)+(√2))/((√5)−(√2)))).](Q63767.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\sqrt{\mathrm{3}}{cosx}\:+\sqrt{\mathrm{2}}{sinx}}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left\{\sqrt{\mathrm{3}}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\sqrt{\mathrm{2}}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right\}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\sqrt{\mathrm{3}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{2}\sqrt{\mathrm{2}}{t}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{−\sqrt{\mathrm{3}}{t}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{t}\:+\sqrt{\mathrm{3}}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{−\mathrm{2}{dt}}{\sqrt{\mathrm{3}}{t}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{t}−\sqrt{\mathrm{3}}} \\ $$$$\Delta^{'} =\left(−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{3}\:=\mathrm{5}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\sqrt{\mathrm{3}}}\:{and}\:{t}_{\mathrm{2}} =\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{5}}}{\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${A}\:=\frac{−\mathrm{2}}{\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}\:=\frac{−\mathrm{2}}{\sqrt{\mathrm{3}}\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\:\int_{\mathrm{0}} ^{\infty} \:\left\{\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}{dt} \\ $$$$=\frac{−\mathrm{2}}{\sqrt{\mathrm{3}}\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\sqrt{\mathrm{3}}}}\:\int_{\mathrm{0}} ^{\infty} \:\left\{\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right\}{dt}=−\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left[{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}{ln}\mid\frac{{t}_{\mathrm{1}} }{{t}_{\mathrm{2}} }\mid\:=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}{ln}\mid\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\sqrt{\mathrm{2}}−\sqrt{\mathrm{5}}}\mid\:=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}{ln}\left(\frac{\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}}{\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}}\right). \\ $$
Answered by MJS last updated on 07/Jul/19
$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\int\frac{{dx}}{\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)} \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{again}... \\ $$
Commented by mathmax by abdo last updated on 07/Jul/19
$${thank}\:{you}\:{sir}. \\ $$