find the value of ∫_0 ^(π/2) (x/(√(1−cosx)))dx.


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Question Number 43999 by maxmathsup by imad last updated on 19/Sep/18

$f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{x}{\sqrt{1 - c o s x}} d x .$
Commented by maxmathsup by imad last updated on 20/Sep/18
$let I = ∫_0 ^(π/2) (x/(√(1−cosx)))dx ⇒I =∫_0 ^(π/2) (x/(√(2sin^2 ((x/2)))))dx =(1/(√2)) ∫_0 ^(π/2) (x/(sin((x/2))))dx =_((x/2)=t) (1/(√2)) ∫_0 ^(π/4) ((2t)/(sin(t))) 2 dt =(4/(√2)) ∫_0 ^(π/4) (t/(sint))dt changement tan((t/2))=u give I = (4/(√2)) ∫_0 ^((√2)−1) ((2arctan(u))/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) =4(√( 2)) ∫_0 ^((√2)−1) ((arctan(u))/u) du let f(x)= ∫_0 ^((√2)−1) ((arctan(xu))/u) du ⇒ I =4(√2) f(1) we have f^′ (x) = ∫_0 ^((√2)−1) (du/(1+x^2 u^2 )) =_(xu =α) ∫_0 ^(x((√2)−1)) (1/(1+α^2 )) (dα/x) = (1/x) ∫_0 ^(x((√2)−1)) (dα/(1+α^2 )) =(1/x) [arctan(α)]_0 ^(x((√2)−1)) =((arctan{x((√2)−1)})/x) ⇒f(x) =∫_0 ^x ((arctan{t((√2)−1)})/t) dt +k k=f(0) =0 ⇒f(x) =∫_0 ^x ((arctan{t((√2)−1)})/t)dt ⇒ f(1) =∫_0 ^1 ((arctan{t((√2) −1)})/t)dtwe have arctan^′ (x) =(1/(1+x^2 )) =Σ_(n=0) ^∞ (−1)^n x^(2n) ⇒arctanx =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) x^(2n+1) ⇒ arctan{t((√2)−1)} =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) t^(2n+1) ((√2)−1)^(2n+1) ⇒ f(1) =Σ_(n=0) ^∞ ∫_0 ^1 (((−1)^n ((√2)−1)^(2n+1) )/(2n+1)) t^(2n) dt =Σ_(n=0) ^∞ (((−1)^n ((√2)−1)^(2n+1) )/((2n+1)^2 )) and I =4(√2)f(1)...$
$l e t I = \int_{0}^{\frac{π}{2}} \frac{x}{\sqrt{1 - c o s x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{x}{\sqrt{2 {s i n}^{2} (\frac{x}{2})}} d x$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{x}{s i n (\frac{x}{2})} d x =_{\frac{x}{2} = t} \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{4}} \frac{2 t}{s i n (t)} 2 d t = \frac{4}{\sqrt{2}} \int_{0}^{\frac{π}{4}} \frac{t}{s i n t} d t$ $c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $I = \frac{4}{\sqrt{2}} \int_{0}^{\sqrt{2} - 1} \frac{2 a r c t a n (u)}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 4 \sqrt{2} \int_{0}^{\sqrt{2} - 1} \frac{a r c t a n (u)}{u} d u$ $l e t f (x) = \int_{0}^{\sqrt{2} - 1} \frac{a r c t a n (x u)}{u} d u \Rightarrow I = 4 \sqrt{2} f (1)$ $w e h a v e f^{^{'}} (x) = \int_{0}^{\sqrt{2} - 1} \frac{d u}{1 + x^{2} u^{2}} =_{x u = α} \int_{0}^{x (\sqrt{2} - 1)} \frac{1}{1 + α^{2}} \frac{d α}{x}$ $= \frac{1}{x} \int_{0}^{x (\sqrt{2} - 1)} \frac{d α}{1 + α^{2}} = \frac{1}{x} {[a r c t a n (α)]}_{0}^{x (\sqrt{2} - 1)}$ $= \frac{a r c t a n {x (\sqrt{2} - 1)}}{x} \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n {t (\sqrt{2} - 1)}}{t} d t + k$ $k = f (0) = 0 \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n {t (\sqrt{2} - 1)}}{t} d t \Rightarrow$ $f (1) = \int_{0}^{1} \frac{a r c t a n {t (\sqrt{2} - 1)}}{t} d t w e h a v e$ ${a r c t a n}^{^{'}} (x) = \frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} \Rightarrow a r c t a n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} \Rightarrow$ $a r c t a n {t (\sqrt{2} - 1)} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n + 1} {(\sqrt{2} - 1)}^{2 n + 1} \Rightarrow$ $f (1) = \sum_{n = 0}^{\infty} \int_{0}^{1} \frac{{(- 1)}^{n} {(\sqrt{2} - 1)}^{2 n + 1}}{2 n + 1} t^{2 n} d t$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(\sqrt{2} - 1)}^{2 n + 1}}{{(2 n + 1)}^{2}} a n d I = 4 \sqrt{2} f (1) . . .$
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