∫_( 0) ^1 (dx/(x^2 +2x cos α+1)) = α sin α


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Question Number 64139 by Chi Mes Try last updated on 14/Jul/19

$\underset{0}{\int^{1}} \frac{d x}{x^{2} + 2 x \cos α + 1} = α \sin α$
Commented by Prithwish sen last updated on 14/Jul/19

$\int_{0}^{1} \frac{dx}{{(x + \cos α)}^{2} + {(\sin α)}^{2}}$ $= {[\frac{1}{(\sin α)} . \tan^{- 1} (\frac{x + \cos}{\sin α})]}_{0}^{1}$
Commented by mathmax by abdo last updated on 14/Jul/19
$let I =∫_0 ^1 (dx/(x^2 +2xcosα +1)) ⇒I =∫_0 ^1 (dx/(x^2 +2xcosα +cos^2 α +sin^2 α)) =∫_0 ^1 (dx/((x+cosα)^2 +sin^2 α)) changement x+cosα =sinα t give I = ∫_(cotanα) ^((1+cosα)/(sinα)) ((sinα dt)/(sin^2 α(1+t^2 ))) =(1/(sinα))[arctant]_(1/(tanα)) ^((1+cosα)/(sinα)) =(1/(sinα)){ arctan(((1+cosα)/(sinα)))−arctan((1/(tanα)))but ((1+cosα)/(sinα)) =((2cos^2 ((α/2)))/(2sin((α/2))cos((α/2)))) =(1/(tan((α/2)))) ⇒ arctan(((1+cosα)/(sinα)))=+^− (π/2) −(α/2) arctan((1/(tanα)))=+^− (π/2) −α ⇒ I =(1/(sinα)){+^− (π/2) +^− (π/2) +(α/2)} so there is a error in the question if not a equation...$
$l e t I = \int_{0}^{1} \frac{d x}{x^{2} + 2 x c o s α + 1} \Rightarrow I = \int_{0}^{1} \frac{d x}{x^{2} + 2 x c o s α + {c o s}^{2} α + {s i n}^{2} α}$ $= \int_{0}^{1} \frac{d x}{{(x + c o s α)}^{2} + {s i n}^{2} α} c h a n g e m e n t x + c o s α = s i n α t g i v e$ $I = \int_{c o t a n α}^{\frac{1 + c o s α}{s i n α}} \frac{s i n α d t}{{s i n}^{2} α (1 + t^{2})} = \frac{1}{s i n α} {[a r c t a n t]}_{\frac{1}{t a n α}}^{\frac{1 + c o s α}{s i n α}}$ $= \frac{1}{s i n α} {a r c t a n (\frac{1 + c o s α}{s i n α}) - a r c t a n (\frac{1}{t a n α}) b u t$ $\frac{1 + c o s α}{s i n α} = \frac{2 {c o s}^{2} (\frac{α}{2})}{2 s i n (\frac{α}{2}) c o s (\frac{α}{2})} = \frac{1}{t a n (\frac{α}{2})} \Rightarrow$ $a r c t a n (\frac{1 + c o s α}{s i n α}) = \bar{+} \frac{π}{2} - \frac{α}{2}$ $a r c t a n (\frac{1}{t a n α}) = \bar{+} \frac{π}{2} - α \Rightarrow I = \frac{1}{s i n α} {\bar{+} \frac{π}{2} \bar{+} \frac{π}{2} + \frac{α}{2}}$ $s o t h e r e i s a e r r o r i n t h e q u e s t i o n i f n o t a e q u a t i o n . . .$
	Answered by Hope last updated on 15/Jul/19

	$D_{r} = {(x + c o s α)}^{2} + {s i n}^{2} α$ $t = x + c o s α$ $\int_{c o s α}^{1 + c o s α} \frac{d t}{t^{2} + {s i n}^{2} α}$ $\frac{1}{s i n α} ∣ {t a n}^{- 1} (\frac{t}{s i n α}) ∣_{c o s α}^{1 + c o s α}$ $= \frac{1}{s i n α} [{t a n}^{- 1} (\frac{1 + c o s α}{s i n α}) - {t a n}^{- 1} (\frac{c o s α}{s i n α})]$ $= \frac{1}{s i n α} [{t a n}^{- 1} (c o t \frac{α}{2}) - {t a n}^{- 1} (c o t α)]$ $= \frac{1}{s i n α} [{t a n}^{- 1} (t a n (\frac{π}{2} - \frac{α}{2})) - {t a n}^{- 1} (\frac{π}{2} - α)]$ $= \frac{1}{s i n α} [\frac{π}{2} - \frac{α}{2} - \frac{π}{2} + α]$ $= \frac{α}{2 s i n α}$
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