calculate ∫_(−(π/6)) ^(π/6) (x/(sinx))dx


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Question Number 65924 by mathmax by abdo last updated on 05/Aug/19

$c a l c u l a t e \int_{- \frac{π}{6}}^{\frac{π}{6}} \frac{x}{s i n x} d x$
Commented by mathmax by abdo last updated on 07/Aug/19
$let I =∫_(−(π/6)) ^(π/6) (x/(sinx))dx let find approximate value we have I =2∫_0 ^(π/6) (x/(sinx))dx but sinx =Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)!)) with radiusR=+∞ ⇒sinx =x−(x^3 /(3!)) +(x^5 /(5!))−....⇒x−(x^3 /(3!))≤sinx ≤x ⇒(1/x)≤(1/(sinx))≤(1/(x−(x^3 /6))) ⇒1≤(x/(sinx))≤(1/(1−(x^2 /6))) for x∈]0,(π/6)] ⇒∫_0 ^(π/6) 1dx ≤∫_0 ^(π/6) (x/(sinx))dx≤∫_0 ^(π/6) ((6dx)/(6−x^2 )) (π/3)≤ 2∫_0 ^(π/6) (x/(sinx))dx ≤12 ∫_0 ^(π/6) (dx/(6−x^2 )) ⇒(π/3)≤I ≤12∫_0 ^(π/6) (dx/(6−x^2 )) ∫_0 ^(π/6) (dx/(6−x^2 )) =−∫_0 ^(π/6) (dx/((x−(√6))(x+(√6)))) =−(1/(2(√6)))∫_0 ^(π/6) {(1/(x−(√6)))−(1/(x+(√6)))}dx =−(1/(2(√6)))[ln∣((x−(√6))/(x+(√6)))∣]_0 ^(π/6) =−(1/(2(√6)))ln∣(((π/6)−(√6))/((π/6)+(√6)))∣ =−(1/(2(√6)))ln∣((π−6(√6))/(π+6(√6)))∣ =(1/(2(√6)))ln(((π+6(√6))/(6(√6)−π))) ⇒(π/3)≤ I ≤(√6)ln(((6(√6)+π)/(6(√6)−π))) let v_0 =(π/6) +((√6)/2)ln(((6(√6)+π)/(6(√6)−π))) v_0 is a better approximation for I .$
$l e t I = \int_{- \frac{π}{6}}^{\frac{π}{6}} \frac{x}{s i n x} d x l e t f i n d a p p r o x i m a t e v a l u e w e h a v e$ $I = 2 \int_{0}^{\frac{π}{6}} \frac{x}{s i n x} d x b u t s i n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1)!} w i t h r a d i u s R = + \infty$ $\Rightarrow s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . . . \Rightarrow x - \frac{x^{3}}{3!} ⩽ s i n x ⩽ x \Rightarrow \frac{1}{x} ⩽ \frac{1}{s i n x} ⩽ \frac{1}{x - \frac{x^{3}}{6}}$ $\Rightarrow 1 ⩽ \frac{x}{s i n x} ⩽ \frac{1}{1 - \frac{x^{2}}{6}} f o r x \in] 0, \frac{π}{6}] \Rightarrow \int_{0}^{\frac{π}{6}} 1 d x ⩽ \int_{0}^{\frac{π}{6}} \frac{x}{s i n x} d x ⩽ \int_{0}^{\frac{π}{6}} \frac{6 d x}{6 - x^{2}}$ $\frac{π}{3} ⩽ 2 \int_{0}^{\frac{π}{6}} \frac{x}{s i n x} d x ⩽ 12 \int_{0}^{\frac{π}{6}} \frac{d x}{6 - x^{2}} \Rightarrow \frac{π}{3} ⩽ I ⩽ 12 \int_{0}^{\frac{π}{6}} \frac{d x}{6 - x^{2}}$ $\int_{0}^{\frac{π}{6}} \frac{d x}{6 - x^{2}} = - \int_{0}^{\frac{π}{6}} \frac{d x}{(x - \sqrt{6}) (x + \sqrt{6})} = - \frac{1}{2 \sqrt{6}} \int_{0}^{\frac{π}{6}} {\frac{1}{x - \sqrt{6}} - \frac{1}{x + \sqrt{6}}} d x$ $= - \frac{1}{2 \sqrt{6}} {[l n ∣ \frac{x - \sqrt{6}}{x + \sqrt{6}} ∣]}_{0}^{\frac{π}{6}} = - \frac{1}{2 \sqrt{6}} l n ∣ \frac{\frac{π}{6} - \sqrt{6}}{\frac{π}{6} + \sqrt{6}} ∣$ $= - \frac{1}{2 \sqrt{6}} l n ∣ \frac{π - 6 \sqrt{6}}{π + 6 \sqrt{6}} ∣ = \frac{1}{2 \sqrt{6}} l n (\frac{π + 6 \sqrt{6}}{6 \sqrt{6} - π}) \Rightarrow \frac{π}{3} ⩽ I ⩽ \sqrt{6} l n (\frac{6 \sqrt{6} + π}{6 \sqrt{6} - π})$ $l e t v_{0} = \frac{π}{6} + \frac{\sqrt{6}}{2} l n (\frac{6 \sqrt{6} + π}{6 \sqrt{6} - π})$ $v_{0} i s a b e t t e r a p p r o x i m a t i o n f o r I .$
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