calculate ∫_0 ^π (dx/((√3)cosx +(√2)sinx))


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Question Number 63711 by mathmax by abdo last updated on 07/Jul/19

$c a l c u l a t e \int_{0}^{π} \frac{d x}{\sqrt{3} c o s x + \sqrt{2} s i n x}$
Commented by mathmax by abdo last updated on 08/Jul/19
$let A =∫_0 ^π (dx/((√3)cosx +(√2)sinx)) changement tan((x/2))=t give A =∫_0 ^∞ ((2dt)/((1+t^2 ){(√3)((1−t^2 )/(1+t^2 )) +(√2)((2t)/(1+t^2 ))})) =∫_0 ^∞ ((2dt)/((√3)(1−t^2 )+2(√2)t)) =∫_0 ^∞ ((2dt)/(−(√3)t^2 +2(√2)t +(√3))) =∫_0 ^∞ ((−2dt)/((√3)t^2 −2(√2)t−(√3))) Δ^′ =(−(√2))^2 +3 =5 ⇒t_1 =(((√2)+(√5))/(√3)) and t_2 =(((√2)−(√5))/(√3)) ⇒ A =((−2)/(√3))∫_0 ^∞ (dt/((t−t_1 )(t−t_2 ))) =((−2)/((√3)(t_1 −t_2 ))) ∫_0 ^∞ {(1/(t−t_1 )) −(1/(t−t_2 ))}dt =((−2)/((√3)((2(√5))/(√3)))) ∫_0 ^∞ {(1/(t−t_1 )) −(1/(t−t_2 ))}dt=−(1/(√5))[ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^(+∞) =(1/(√5))ln∣(t_1 /t_2 )∣ =(1/(√5))ln∣(((√2)+(√5))/((√2)−(√5)))∣ =(1/(√5))ln((((√5)+(√2))/((√5)−(√2)))).$
$l e t A = \int_{0}^{π} \frac{d x}{\sqrt{3} c o s x + \sqrt{2} s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $A = \int_{0}^{\infty} \frac{2 d t}{(1 + t^{2}) {\sqrt{3} \frac{1 - t^{2}}{1 + t^{2}} + \sqrt{2} \frac{2 t}{1 + t^{2}}}}$ $= \int_{0}^{\infty} \frac{2 d t}{\sqrt{3} (1 - t^{2}) + 2 \sqrt{2} t} = \int_{0}^{\infty} \frac{2 d t}{- \sqrt{3} t^{2} + 2 \sqrt{2} t + \sqrt{3}} = \int_{0}^{\infty} \frac{- 2 d t}{\sqrt{3} t^{2} - 2 \sqrt{2} t - \sqrt{3}}$ $Δ^{^{'}} = {(- \sqrt{2})}^{2} + 3 = 5 \Rightarrow t_{1} = \frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}} a n d t_{2} = \frac{\sqrt{2} - \sqrt{5}}{\sqrt{3}} \Rightarrow$ $A = \frac{- 2}{\sqrt{3}} \int_{0}^{\infty} \frac{d t}{(t - t_{1}) (t - t_{2})} = \frac{- 2}{\sqrt{3} (t_{1} - t_{2})} \int_{0}^{\infty} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} d t$ $= \frac{- 2}{\sqrt{3} \frac{2 \sqrt{5}}{\sqrt{3}}} \int_{0}^{\infty} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} d t = - \frac{1}{\sqrt{5}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{+ \infty}$ $= \frac{1}{\sqrt{5}} l n ∣ \frac{t_{1}}{t_{2}} ∣ = \frac{1}{\sqrt{5}} l n ∣ \frac{\sqrt{2} + \sqrt{5}}{\sqrt{2} - \sqrt{5}} ∣ = \frac{1}{\sqrt{5}} l n (\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}) .$
	Answered by MJS last updated on 07/Jul/19

	$= \frac{\sqrt{5}}{5} \int \frac{d x}{\sin (x + \arctan \frac{\sqrt{6}}{2})}$ $now {it}^{'} s easy again . . .$
	Commented by mathmax by abdo last updated on 07/Jul/19

	$t h a n k y o u s i r .$
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