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


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

$Thr value of$ $\underset{0}{\int^{π / 2}} cosec (x - \frac{π}{3}) cosec (x - \frac{π}{6}) d x is$
Commented by mathmax by abdo last updated on 21/Dec/19
$I =∫_0 ^(π/2) (dx/(sin(x−(π/3))sin(x−(π/6)))) but we have sin(x−(π/3))sin(x−(π/6)) =cos((π/2)−x+(π/3))cos((π/2)−x +(π/6)) =cos(((5π)/6)−x)cos(((2π)/3)−x) =(1/2){ cos(((5π)/6)+((2π)/3)−2x) +cos(((5π)/6)−x−((2π)/3)+x)} =(1/2){cos(((3π)/2)−2x) +cos((π/6))}=(1/2){cos(−(π/2)−2x)+((√3)/2)} =(1/2){((√3)/2)−sin(2x)} ⇒I =2 ∫_0 ^(π/2) (dx/(((√3)/2)−sin(2x))) =4 ∫_0 ^(π/2) (dx/((√3)−2sin(2x))) =_(2x=t) 4 ∫_0 ^π (dt/(2((√3)−2sint))) =2 ∫_0 ^π (dt/((√3)−2sin(t))) =_(tan((t/2))=u) 2∫_0 ^(+∞) (1/((√3)−2((2u)/(1+u^2 ))))((2du)/(1+u^2 )) =4 ∫_0 ^∞ (du/((√3)+(√3)u^2 −4u)) =4 ∫_0 ^∞ (du/((√3)u^2 −4u +(√3))) Δ^′ =(−2)^2 −3 =1 ⇒u_1 =((2+1)/(√3)) =(3/(√3)) =(√3) u_2 =((2−1)/(√3)) =(1/(√3)) ⇒I =4 ∫_0 ^∞ (du/((√3)(u−(√3))(u−(1/(√3))))) =(4/(√3))×(1/((√3)−(1/(√3))))∫_0 ^∞ ((1/(u−(√3)))−(1/(u−(1/(√3)))))du =2 ln∣((u−(√3))/(u−(1/(√3))))∣ +c =2ln∣((tan(x)−(√3))/(tan(x)−(1/(√3))))∣ +c$
$I = \int_{0}^{\frac{π}{2}} \frac{d x}{s i n (x - \frac{π}{3}) s i n (x - \frac{π}{6})} b u t w e h a v e$ $s i n (x - \frac{π}{3}) s i n (x - \frac{π}{6}) = c o s (\frac{π}{2} - x + \frac{π}{3}) c o s (\frac{π}{2} - x + \frac{π}{6})$ $= c o s (\frac{5 π}{6} - x) c o s (\frac{2 π}{3} - x)$ $= \frac{1}{2} {c o s (\frac{5 π}{6} + \frac{2 π}{3} - 2 x) + c o s (\frac{5 π}{6} - x - \frac{2 π}{3} + x)}$ $= \frac{1}{2} {c o s (\frac{3 π}{2} - 2 x) + c o s (\frac{π}{6})} = \frac{1}{2} {c o s (- \frac{π}{2} - 2 x) + \frac{\sqrt{3}}{2}}$ $= \frac{1}{2} {\frac{\sqrt{3}}{2} - s i n (2 x)} \Rightarrow I = 2 \int_{0}^{\frac{π}{2}} \frac{d x}{\frac{\sqrt{3}}{2} - s i n (2 x)}$ $= 4 \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{3} - 2 s i n (2 x)} =_{2 x = t} 4 \int_{0}^{π} \frac{d t}{2 (\sqrt{3} - 2 s i n t)}$ $= 2 \int_{0}^{π} \frac{d t}{\sqrt{3} - 2 s i n (t)} =_{t a n (\frac{t}{2}) = u} 2 \int_{0}^{+ \infty} \frac{1}{\sqrt{3} - 2 \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= 4 \int_{0}^{\infty} \frac{d u}{\sqrt{3} + \sqrt{3} u^{2} - 4 u} = 4 \int_{0}^{\infty} \frac{d u}{\sqrt{3} u^{2} - 4 u + \sqrt{3}}$ $Δ^{^{'}} = {(- 2)}^{2} - 3 = 1 \Rightarrow u_{1} = \frac{2 + 1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ $u_{2} = \frac{2 - 1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow I = 4 \int_{0}^{\infty} \frac{d u}{\sqrt{3} (u - \sqrt{3}) (u - \frac{1}{\sqrt{3}})}$ $= \frac{4}{\sqrt{3}} \times \frac{1}{\sqrt{3} - \frac{1}{\sqrt{3}}} \int_{0}^{\infty} (\frac{1}{u - \sqrt{3}} - \frac{1}{u - \frac{1}{\sqrt{3}}}) d u$ $= 2 l n ∣ \frac{u - \sqrt{3}}{u - \frac{1}{\sqrt{3}}} ∣ + c$ $= 2 l n ∣ \frac{t a n (x) - \sqrt{3}}{t a n (x) - \frac{1}{\sqrt{3}}} ∣ + c$
Commented by benjo last updated on 22/Dec/19

$thanks sir$
Commented by abdomathmax last updated on 23/Dec/19

$y o u a r e w e l c o m e .$
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