let f(x)=∫_0 ^(π/2) ln(1+xcosθ)dθ 1) calculate f(1) 2) find a simple form of f(x) 3) developp f at ontehr serie


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Question Number 40621 by math khazana by abdo last updated on 25/Jul/18

$l e t f (x) = \int_{0}^{\frac{π}{2}} l n (1 + x c o s θ) d θ$ $1) c a l c u l a t e f (1)$ $2) f i n d a s i m p l e f o r m o f f (x)$ $3) d e v e l o p p f a t o n t e h r s e r i e$
	Answered by math khazana by abdo last updated on 27/Jul/18

	$1) f (1) = \int_{0}^{\frac{π}{2}} l n (1 + c o s θ) d θ = I a n d l e t$ $J = \int_{0}^{\frac{π}{2}} l n (1 - c o s θ) d θ w e h a v e$ $I + J = \int_{0}^{\frac{π}{2}} l n (1 - {c o s}^{2} θ) d θ = \int_{0}^{\frac{π}{2}} l n ({s i n}^{2} θ) d θ$ $= 2 \int_{0}^{\frac{π}{2}} l n (s i n θ) d θ = 2 (- \frac{π}{2} l n (2)) = - π l n (2)$ $I - J = \int_{0}^{\frac{π}{2}} l n (\frac{1 + c o s θ}{1 - c o s θ}) d θ$ $= \int_{0}^{\frac{π}{2}} l n (\frac{2 {c o s}^{2} (\frac{θ}{2})}{2 {s i n}^{2} (\frac{θ}{2})}) d θ = \int_{0}^{\frac{π}{2}} - 2 l n (t a n (\frac{θ}{2})) d θ$ $= - 2 \int_{0}^{\frac{π}{2}} l n (t a n (\frac{θ}{2})) d θ$ $=_{\frac{θ}{2} = t} - 4 \int_{0}^{\frac{π}{4}} l n (t a n (t)) d t$ $\int_{0}^{\frac{π}{4}} l n (t a n t) d t =_{t a n t = u} \int_{0}^{1} l n (u) \frac{d u}{1 + u^{2}}$ $= \int_{0}^{1} \frac{l n (u)}{1 + u^{2}} d u = \int_{0}^{1} l n (u) \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{2 n}) d u$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} u^{2 n} l n (u) d u = \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n}$ $A_{n} = {[\frac{1}{2 n + 1} u^{2 n + 1} l n (u)]}_{0}^{1} - \int_{0}^{1} \frac{u^{2 n}}{(2 n + 1)} d u$ $= - \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \int_{0}^{\frac{π}{4}} l n (t a n t) d t = - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}}$ $= - λ_{0} (t h e v a l u e o f λ_{0} i s k n o w n) \Rightarrow$ $I - J = 4 λ_{0} \Rightarrow I + J = - π l n (2) a n d$ $I - J = 4 λ_{0} \Rightarrow 2 I = - π l n (2) + 4 λ_{0} \Rightarrow$ $I = - \frac{π}{2} l n (2) + 2 λ_{0}$
	Answered by math khazana by abdo last updated on 27/Jul/18

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