let f(x) =∫_0 ^(π/2) ln(1−xcost)dt with ∣x∣<1 1) developp f at integr serie 2) find a explicit form of f(x) 3) find the values of integrals ∫_0 ^(π/2) ln(1−cost)dt and ∫_0 ^(π/2) ln(1+cost)dt 4) calculate U_n =∫_0 ^(π/2) ln(1−(2/n)cost)dt with n integr and n≥2 study the convergence of U_n and Σ U


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Question Number 59247 by maxmathsup by imad last updated on 06/May/19

$l e t f (x) = \int_{0}^{\frac{π}{2}} l n (1 - x c o s t) d t w i t h ∣ x ∣< 1$ $1) d e v e l o p p f a t i n t e g r s e r i e$ $2) f i n d a e x p l i c i t f o r m o f f (x)$ $3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\frac{π}{2}} l n (1 - c o s t) d t a n d \int_{0}^{\frac{π}{2}} l n (1 + c o s t) d t$ $4) c a l c u l a t e U_{n} = \int_{0}^{\frac{π}{2}} l n (1 - \frac{2}{n} c o s t) d t w i t h n i n t e g r a n d n ⩾ 2$ $s t u d y t h e c o n v e r g e n c e o f U_{n} a n d Σ U_{n}$
Commented byMr X pcx last updated on 09/May/19

$2) w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{- c o s t}{1 - x c o s t} d t$ $f o r x \neq 0 f^{^{'}} (x) = \frac{1}{x} \int_{0}^{\frac{π}{2}} \frac{1 - x c o s t - 1}{1 - x c o s t} d t$ $= \frac{π}{2 x} - \frac{1}{x} \int_{0}^{\frac{π}{2}} \frac{d t}{1 - x c o s t} b u t$ $c h a n g . t a n (\frac{t}{2}) = u g i v e$ $\int_{0}^{\frac{π}{2}} \frac{d t}{1 - x c o s t} = \int_{0}^{1} \frac{1}{1 - x \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \int_{0}^{1} \frac{2 d u}{1 + u^{2} - x + {x u}^{2}} = \int_{0}^{1} \frac{2 d u}{(1 + x) u^{2} + 1 - x}$ $= \frac{2}{1 + x} \int_{0}^{1} \frac{d u}{u^{2} + \frac{1 - x}{1 + x}}$ $=_{u = \sqrt{\frac{1 - x}{1 + x}} α} \frac{2}{1 + x} \int_{0}^{\sqrt{\frac{1 + x}{1 - x}}} \frac{1}{\frac{1 - x}{1 + x} (1 + α^{2})} \sqrt{\frac{1 - x}{1 + x}} d α$ $= \frac{2}{1 - x} \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \int_{0}^{\sqrt{\frac{1 + x}{1 - x}}} \frac{d α}{1 + α^{2}}$ $= \frac{2}{\sqrt{1 - x^{2}}} {[a r c t a n (α)]}_{0}^{\sqrt{\frac{1 + x}{1 - x}}}$ $= \frac{2}{\sqrt{1 - x^{2}}} a r c t a n (\sqrt{\frac{1 + x}{1 - x}}) \Rightarrow$ $f^{^{'}} (x) = \frac{π}{2 x} - \frac{2}{x \sqrt{1 - x^{2}}} a r c t a n (\sqrt{\frac{1 + x}{1 - x}}) \Rightarrow$ $f (x) = \frac{π}{2} l n ∣ x ∣ - \int \frac{2}{x \sqrt{1 - x^{2}}} a r c t a n (\sqrt{\frac{1 + x}{1 - x}}) d x + c$ $\int \frac{2}{x \sqrt{1 - x^{2}}} a r c t a n (\sqrt{\frac{1 + x}{1 - x}}) d x$ $=_{x = c o s t} \int \frac{2}{c o s t . s i n t} a r c t a n (\sqrt{\frac{2 {c o s}^{2} (\frac{t}{2})}{2 {s i n}^{2} (\frac{t}{2})}}) (- s i n t) d t$ $= - 2 \int \frac{1}{s i n t} a r c t a n (\frac{1}{t a n (\frac{t}{2})}) d t$ $= - 2 \int \frac{1}{s i n t} (\frac{π}{2} - \frac{t}{2}) d t$ $= - π \int \frac{d t}{s i n t} + \int \frac{t}{s i n t} d t . . . . b e c o n t i n u e d . . .$
Commented bymaxmathsup by imad last updated on 09/May/19

$e r r o r a t l i n e 16 w e h a v e \int \frac{2}{x \sqrt{1 - x^{2}}} a r c t a n (\sqrt{\frac{1 + x}{1 - x}}) d x$ $= - 2 \int \frac{1}{c o s t} (\frac{π}{2} - \frac{t}{2}) d t + c = - π \int \frac{d t}{c o s t} + \int \frac{t}{c o s t} d t + c \Rightarrow$ $f (x) = \frac{π}{2} l n ∣ x ∣ + π \int \frac{d t}{c o s t} - \int \frac{t}{c o s t} d t + c . . . .$
	Answered by Mr X pcx last updated on 09/May/19

	$w e h a v e {l n}^{^{'}} (1 - u) = \frac{- 1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n}$ $w i t h ∣ u ∣< 1 \Rightarrow l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1}$ $= - \sum_{n = 1}^{\infty} \frac{u^{n}}{n} w e h a v e ∣ x c o s t ∣< 1 \Rightarrow$ $l n (1 - x c o s t) = - \sum_{n = 1}^{\infty} \frac{x^{n} {c o s}^{n} t}{n} \Rightarrow$ $\int_{0}^{\frac{π}{2}} l n (1 - x c o s t) d t = - \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \int_{0}^{\frac{π}{2}} {c o s}^{n} (t)$ $= - \sum_{n = 1}^{\infty} \frac{A_{n}}{n} x^{n} {w i t h A}_{n} = \int_{0}^{\frac{π}{2}} {c o s}^{n} t d t$ $(i n t e g r a l o f w a l l i s) .$
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