calculate ∫_(π/8) ^(π/6) (dx/(sin(2x)))


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Question Number 36427 by prof Abdo imad last updated on 02/Jun/18

$c a l c u l a t e \int_{\frac{π}{8}}^{\frac{π}{6}} \frac{d x}{s i n (2 x)}$
Commented by abdo mathsup 649 cc last updated on 03/Jun/18
$I =_(2x =u) ∫_(π/4) ^(π/3) (1/(sin(u))) (du/2) = (1/2) ∫_(π/4) ^(π/3) (du/(sinu)) and changement tan((u/2))=x give I = (1/2) ∫_((√2) −1) ^(1/(√3)) (1/((2x)/(1+x^2 ))) ((2dx)/(1+x^2 )) = (1/2) ∫_((√2)−1) ^(1/(√3)) (dx/x) =(1/2)[ln ∣x∣]_((√2) −1) ^(1/(√3)) = (1/2){ −ln((√3)) −ln((√2) −1) =−(1/2){ (1/2)ln(3) +ln((√2) −1)} .$
$I =_{2 x = u} \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{s i n (u)} \frac{d u}{2}$ $= \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d u}{s i n u} a n d c h a n g e m e n t t a n (\frac{u}{2}) = x g i v e$ $I = \frac{1}{2} \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{1}{\frac{2 x}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = \frac{1}{2} \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{d x}{x}$ $= \frac{1}{2} {[l n ∣ x ∣]}_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} = \frac{1}{2} {- l n (\sqrt{3}) - l n (\sqrt{2} - 1)$ $= - \frac{1}{2} {\frac{1}{2} l n (3) + l n (\sqrt{2} - 1)} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
	$∫_(Π/8) ^(Π/6) ((1+tan^2 x)/(2tanx))dx (1/2)∣ln(tanx)∣_(Π/8) ^(Π/6) =(1/2){lntan(Π/6)−lntan(Π/8))$
	$\int_{\frac{Π}{8}}^{\frac{Π}{6}} \frac{1 + {t a n}^{2} x}{2 t a n x} d x$ $\frac{1}{2} ∣ l n (t a n x) ∣_{\frac{Π}{8}}^{\frac{Π}{6}}$ $= \frac{1}{2} {l n t a n \frac{Π}{6} - l n t a n \frac{Π}{8})$
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