∫ _0 ^(2π) (dx/((√2)−cos x))


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Question Number 85568 by jagoll last updated on 23/Mar/20

$\int \underset{0}{\overset{2 π}{}} \frac{dx}{\sqrt{2} - \cos x}$
Commented by jagoll last updated on 23/Mar/20

$I = \int \underset{0}{\overset{2 π}{}} \frac{dx}{\sqrt{2} - \cos x}$ $let x = 2 π - t$ $I = \int \underset{2 π}{\overset{0}{}} \frac{- dt}{\sqrt{2} - \cos (2 π - t)}$ $I = \int \underset{0}{\overset{2 π}{}} \frac{dt}{\sqrt{2} - \cos t}$ $I = I (not valid)$
Commented by mathmax by abdo last updated on 23/Mar/20

$A = \int_{0}^{2 π} \frac{d x}{\sqrt{2} - c o s x} \Rightarrow A = \int_{0}^{2 π} \frac{d x}{\sqrt{2} - \frac{e^{i x} + e^{- i x}}{2}}$ $= \int_{0}^{2 π} \frac{2 d x}{2 \sqrt{2} - e^{i x} - e^{- i x}} =_{e^{i x} = z} \int_{∣ z ∣= 1} \frac{2}{2 \sqrt{2} - z - z^{- 1}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{- 2 i d z}{2 \sqrt{2} z - z^{2} - 1} = \int_{∣ z ∣= 1} \frac{2 i d z}{z^{2} - 2 \sqrt{2} z + 1} = \int_{∣ z ∣= 1} φ (z) d z$ $w i t h φ (z) = \frac{2 i}{z^{2} - 2 \sqrt{2} z + 1} p o l e s o f φ ?$ $z^{2} - 2 \sqrt{2} z + 1 = 0 \to Δ^{^{'}} = 2 - 1 = 1 \Rightarrow z_{1} = 1 + \sqrt{2} a n d z_{2} = - 1 + \sqrt{2}$ $\Rightarrow φ (z) = \frac{2 i}{(z - z_{1}) (z - z_{2})}$ $∣ z_{1} ∣ - 1 = 1 + \sqrt{2} - 1 > 0 (o u t o f c i r c l e)$ $∣ z_{2} ∣ - 1 =∣ \sqrt{2} - 1 ∣ - 1 = \sqrt{2} - 2 < 0 \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{2}) = 2 i π \times \frac{2 i}{(z_{2} - z_{1})} = \frac{- 4 π}{- 2} = 2 π$ $\Rightarrow A = 2 π$
	Answered by mind is power last updated on 23/Mar/20
	$=∫_0 ^π (dx/((√2)−cos(x)))+∫_π ^(2π) (dx/((√2)−cos(x))) t=π+x in 2nd⇒ =∫_0 ^π (dx/((√2)−cos(x)))+(dx/((√2)+cos(x))) =∫_0 ^π ((2(√2)dx)/(2−cos^2 (x)))=2(√2)∫_0 ^(π/2) (dx/(2−cos^2 (x)))+2(√2)∫_(π/2) ^π (dx/(2−cos^2 (x))) t=(π/2)+x in 2nd ⇒=2(√(2{))∫_0 ^(π/2) (dx/(1+sin^2 (x)))+∫_0 ^(π/2) (dx/(1+cos^2 ((x/2)))) =2(√2)∫_0 ^(π/2) (1/(sin^2 (x))).((dx/((1/(sin^2 (x)))+1)))+2(√2)∫_0 ^(π/2) (dx/(cos^2 (x))).(1/(((1/(cos^2 (x)))+1))) =2(√2)∫_0 ^(π/2) ((−d(cot(x)))/(2+cot^2 (x)))+2(√2)∫_0 ^(π/2) ((d(tg(x)))/(2+tg^2 (x))) =2([_0 ^(π/2) −arctan(((cot(x))/(√2)))+arctan(((tg(x))/(√2)))) =2.(π/2)+2.(π/2)=2π$
	$= \int_{0}^{π} \frac{d x}{\sqrt{2} - c o s (x)} + \int_{π}^{2 π} \frac{d x}{\sqrt{2} - c o s (x)}$ $t = π + x i n 2 n d \Rightarrow$ $= \int_{0}^{π} \frac{d x}{\sqrt{2} - c o s (x)} + \frac{d x}{\sqrt{2} + c o s (x)}$ $= \int_{0}^{π} \frac{2 \sqrt{2} d x}{2 - {c o s}^{2} (x)} = 2 \sqrt{2} \int_{0}^{\frac{π}{2}} \frac{d x}{2 - {c o s}^{2} (x)} + 2 \sqrt{2} \int_{\frac{π}{2}}^{π} \frac{d x}{2 - {c o s}^{2} (x)}$ $t = \frac{π}{2} + x i n 2 n d$ $\Rightarrow= 2 \sqrt{2 {} \int_{0}^{\frac{π}{2}} \frac{d x}{1 + {s i n}^{2} (x)} + \int_{0}^{\frac{π}{2}} \frac{d x}{1 + {c o s}^{2} (\frac{x}{2})}$ $= 2 \sqrt{2} \int_{0}^{\frac{π}{2}} \frac{1}{{s i n}^{2} (x)} . (\frac{d x}{\frac{1}{{s i n}^{2} (x)} + 1}) + 2 \sqrt{2} \int_{0}^{\frac{π}{2}} \frac{d x}{{c o s}^{2} (x)} . \frac{1}{(\frac{1}{{c o s}^{2} (x)} + 1)}$ $= 2 \sqrt{2} \int_{0}^{\frac{π}{2}} \frac{- d (c o t (x))}{2 + {c o t}^{2} (x)} + 2 \sqrt{2} \int_{0}^{\frac{π}{2}} \frac{d (t g (x))}{2 + {t g}^{2} (x)}$ $= 2 ([_{0}^{\frac{π}{2}} - a r c t a n (\frac{c o t (x)}{\sqrt{2}}) + a r c t a n (\frac{t g (x)}{\sqrt{2}}))$ $= 2 . \frac{π}{2} + 2 . \frac{π}{2} = 2 π$
	Commented by jagoll last updated on 23/Mar/20

	$sir if by King formula valid ?$
	Commented by mind is power last updated on 23/Mar/20

	$o n f r e n c h i {d i d n}^{'} t r e m e m b e r s t u d i e d$ $w h a t i s k i n g f o r m u l a ?$
	Answered by mind is power last updated on 23/Mar/20

	$= \int_{0}^{2 π} \frac{2 d x}{2 \sqrt{2} - (e^{i x} + e^{- i x})}$ $2 n d W a y$ $= \int_{0}^{2 π} \frac{2 e^{i x}}{- e^{2 i x} + 2 \sqrt{2} e^{i x} - 1}$ $e^{i x} = z \Rightarrow$ $= \int_{C} \frac{2 i d z}{(z - (\sqrt{2} - 1)) (z - (\sqrt{2} + 1))}$ $o n l y \sqrt{2} - 1 \in D (0, 1)$ $= 2 i π . \lim_{z \to \sqrt{2} - 1} (z - (\sqrt{2} - 1)) \frac{2 i}{(z - (\sqrt{2} - 1)) (z - (\sqrt{2} + 1))}$ $= 2 i π . \frac{2 i}{- 2} = 2 π$
	Answered by mind is power last updated on 23/Mar/20

	$y o u c a n a l w a y s d o l v e \int \frac{p (s i n (t), c o s (t))}{Q (s i n (t), c o s (t))} d t$ $w i t h e P s n d Q \in R [x] [y]$ $b y t g (\frac{t}{2}) = x$
	Commented by jagoll last updated on 23/Mar/20

	$oo yes weirstrass sir$
	Answered by MJS last updated on 23/Mar/20

	$\int \frac{d x}{\sqrt{2} - \cos x} =$ $[t = \tan \frac{x}{2} \to d x = \frac{2}{t^{2} + 1} d t]$ $= - 2 (1 - \sqrt{2}) \int \frac{d t}{t^{2} + 3 - 2 \sqrt{2}} =$ $= 2 \arctan ((1 + \sqrt{2}) t) =$ $= 2 \arctan ((1 + \sqrt{2}) \tan \frac{x}{2}) + C$
	Commented by jagoll last updated on 23/Mar/20

	$sir if x = 0 \Rightarrow t = 0$ $if x = 2 π \Rightarrow t = 0$ $does mean : 2 \sqrt{2} - 1 \int_{0}^{0} \frac{dt}{t^{2} + 3 - 2 \sqrt{2}}$ $= 0 ?$ $what wrong sir ?$
	Commented by MJS last updated on 23/Mar/20

	$the given function is symmetric about x = π$ $\Rightarrow \underset{0}{\int^{2 π}} f (x) d x = 2 \underset{0}{\int^{π}} f (x) d x$
	Commented by MJS last updated on 23/Mar/20

	$the borders for t are [0; + \infty [$
	Commented by jagoll last updated on 23/Mar/20

	$Missing \left or extra \right$ $4 \times [arc \tan ((1 + \sqrt{2}) \tan \frac{π}{4})]$ $4 \times [arc \tan ((1 + \sqrt{2}))] ?$
	Commented by jagoll last updated on 23/Mar/20

	$please sir . completely your answer$ $sir$
	Commented by jagoll last updated on 23/Mar/20

	$Missing \left or extra \right$ $4 \times [arc \tan (\infty) - arc \tan (0)]$ $4 \times \frac{π}{2} = 2 π$
	Commented by MJS last updated on 23/Mar/20

	$\underset{0}{\int^{2 π}} \frac{d x}{\sqrt{2} - \cos x} = 2 \underset{0}{\int^{π}} \frac{d x}{\sqrt{2} - \cos x} =$ $= 2 {[2 \arctan ((1 + \sqrt{2}) \tan \frac{x}{2})]}_{0}^{π} =$ $= 4 \lim_{x \to π^{-}} (\arctan ((1 + \sqrt{2}) \tan \frac{x}{2})) = 4 \times \frac{π}{2} = 2 π$
	Commented by jagoll last updated on 23/Mar/20

	$thank you sir$
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