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Question Number 167400 by infinityaction last updated on 15/Mar/22

	Answered by Mathspace last updated on 16/Mar/22

	$I = \frac{π}{2} l n 2 + \int_{0}^{\frac{π}{2}} l n (1 + \frac{1}{2} c o s x) d x$ $= \frac{π}{2} l n 2 + f (\frac{1}{2}) w i t h f (a) = \int_{0}^{\frac{π}{2}} l n (1 + a c o s x) d x$ $w e t a k e 0 < a ⩽ 1$ $f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + a c o s x}$ $= \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{1 + a c o s x - 1}{1 + a c o s x} d x$ $= \frac{π}{2 a} - \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{d x}{1 + a c o s x} (\to t a n (\frac{x}{2}) = t)$ $= \frac{π}{2 a} - \frac{1}{a} \int_{0}^{1} \frac{2 d t}{(1 + t^{2}) (1 + a \frac{1 - t^{2}}{1 + t^{2}})}$ $= \frac{π}{2 a} - \frac{2}{a} \int_{0}^{1} \frac{d t}{1 + t^{2} + a - {a t}^{2}}$ $= \frac{π}{2 a} - \frac{2}{a} \int_{0}^{1} \frac{d t}{1 + a + (1 - a) t^{2}}$ $= \frac{π}{2 a} - \frac{2}{a (1 + a)} \int_{0}^{1} \frac{d t}{1 + \frac{1 - a}{1 + a} t^{2}} (\to \sqrt{\frac{1 - a}{1 + a}} t = y)$ $= \frac{π}{2 a} - \frac{2}{a (1 + a)} \int_{0}^{\sqrt{\frac{1 - a}{1 + a}}} \frac{\sqrt{\frac{1 + a}{1 - a}} d y}{1 + y^{2}}$ $= \frac{π}{2 a} - \frac{2}{a \sqrt{1 - a^{2}}} a r c t a n (\sqrt{\frac{1 - a}{1 + a}}) \Rightarrow$ $f (a) = \frac{π}{2} l n a - 2 \int \frac{a r c t a n (\sqrt{\frac{1 - a}{1 + a}})}{a \sqrt{1 - a^{2}}} d a + C$ $c h a n g e m e n t a = c o s θ g i v e$ $\int \frac{a r c t a n (\sqrt{\frac{1 - a}{1 + a}})}{a \sqrt{1 - a^{2}}} d a$ $= \int \frac{a r c t a n (t a n (\frac{θ}{2})}{c o s θ . s i n θ} (- s i n θ) d θ$ $= - \int \frac{θ}{2 c o s θ} d θ$ $. . . b e c o n t i n u e d . . .$
	Commented by infinityaction last updated on 16/Mar/22

	$t h a n k y o u s i r$
	Answered by mindispower last updated on 16/Mar/22
	$2+cos(x)=(1/2)(4+e^(ix) +e^(−ix) ) =(1/(2e^(ix) ))(e^(2ix) +4e^(ix) +1) a=−2−(√3),b=−2+(√3) =(1/(2e^(ix) ))(e^(ix) −a)(e^(ix) −b) ⇔∫_0 ^(π/2) ln((((e^(ix) −a))/(2e^(ix) )).(e^(ix) −b))dx =(π/2)ln(2)−i((π^2 /8))+∫_0 ^(π/2) ln(e^(ix) −a)dx+∫_0 ^(π/2) ln(e^(ix) −b)dx A=∫_0 ^(π/2) ln(e^(ix) −a)dx =(π/2)ln(−a)+Σ_(k≥1) (1/k)∫_0 ^(π/2) (−((e^(ix) /a))^k )dx =(π/2)(ln(−a))−Σ_(k≥1) (1/(ka^k ))((((i)^k −1)/(ki))) =(π/2)ln(−a)+iΣ_(k≥1) {(((((i/a))^k )/k^2 ))−(((((1/a))^k )/k^2 ))} =(π/2)ln(−a)+i(Li_2 ((i/a))−Li_2 ((1/a))) B=∫_0 ^(π/2) ln(e^(ix) −b)=∫_0 ^(π/2) ixdx+∫_0 ^(π/2) ln(1−be^(−ix) )dx =i((π^2 /8))−Σ_(k≥1) ∫_0 ^(π/2) (((be^(−ix) )^k )/k)dx =i(π^2 /8)−Σ_(k≥1) ((b^k ((−i)^k −1))/(−ik^2 )) =i(π^2 /8)−iΣ_(k≥1) (((−ib)^k −(b)^k )/k^2 ) =((iπ^2 )/8)−iLi_2 (−ib)+iLi_2 (b) ∫_0 ^(π/2) ln(2+cos(x))dx=(π/2)ln(2)+(π/2)ln(−a)+i(Li_2 ((i/a))−Li_2 (−ib)) +i(Li_2 (b)−Li_2 ((1/a))) (1/a)=b,(i/a)=−ib =(π/2)ln(−2a)+i(Li_2 (ib)−Li_2 (−ib)) =(π/2)ln(−2a)−2Im(Li_2 (ib)) =(π/2)ln(4+2(√3))−2Im(Li_2 (i(−2+(√3)))$
	$2 + c o s (x) = \frac{1}{2} (4 + e^{i x} + e^{- i x})$ $= \frac{1}{2 e^{i x}} (e^{2 i x} + 4 e^{i x} + 1)$ $a = - 2 - \sqrt{3}, b = - 2 + \sqrt{3}$ $= \frac{1}{2 e^{i x}} (e^{i x} - a) (e^{i x} - b)$ $\Leftrightarrow \int_{0}^{\frac{π}{2}} l n (\frac{(e^{i x} - a)}{2 e^{i x}} . (e^{i x} - b)) d x$ $= \frac{π}{2} l n (2) - i (\frac{π^{2}}{8}) + \int_{0}^{\frac{π}{2}} l n (e^{i x} - a) d x + \int_{0}^{\frac{π}{2}} l n (e^{i x} - b) d x$ $A = \int_{0}^{\frac{π}{2}} l n (e^{i x} - a) d x$ $= \frac{π}{2} l n (- a) + \sum_{k ⩾ 1} \frac{1}{k} \int_{0}^{\frac{π}{2}} (- {(\frac{e^{i x}}{a})}^{k}) d x$ $= \frac{π}{2} (l n (- a)) - \sum_{k ⩾ 1} \frac{1}{{k a}^{k}} (\frac{{(i)}^{k} - 1}{k i})$ $= \frac{π}{2} l n (- a) + i \sum_{k ⩾ 1} {(\frac{{(\frac{i}{a})}^{k}}{k^{2}}) - (\frac{{(\frac{1}{a})}^{k}}{k^{2}})}$ $= \frac{π}{2} l n (- a) + i ({L i}_{2} (\frac{i}{a}) - {L i}_{2} (\frac{1}{a}))$ $B = \int_{0}^{\frac{π}{2}} l n (e^{i x} - b) = \int_{0}^{\frac{π}{2}} i x d x + \int_{0}^{\frac{π}{2}} l n (1 - {b e}^{- i x}) d x$ $= i (\frac{π^{2}}{8}) - \sum_{k ⩾ 1} \int_{0}^{\frac{π}{2}} \frac{{({b e}^{- i x})}^{k}}{k} d x$ $= i \frac{π^{2}}{8} - \sum_{k ⩾ 1} \frac{b^{k} ({(- i)}^{k} - 1)}{- {i k}^{2}}$ $= i \frac{π^{2}}{8} - i \sum_{k ⩾ 1} \frac{{(- i b)}^{k} - {(b)}^{k}}{k^{2}}$ $= \frac{i π^{2}}{8} - {i L i}_{2} (- i b) + {i L i}_{2} (b)$ $\int_{0}^{\frac{π}{2}} l n (2 + c o s (x)) d x = \frac{π}{2} l n (2) + \frac{π}{2} l n (- a) + i ({L i}_{2} (\frac{i}{a}) - {L i}_{2} (- i b))$ $+ i ({L i}_{2} (b) - {L i}_{2} (\frac{1}{a}))$ $\frac{1}{a} = b, \frac{i}{a} = - i b$ $= \frac{π}{2} l n (- 2 a) + i ({L i}_{2} (i b) - {L i}_{2} (- i b))$ $= \frac{π}{2} l n (- 2 a) - 2 I m ({L i}_{2} (i b))$ $= \frac{π}{2} l n (4 + 2 \sqrt{3}) - 2 I m ({L i}_{2} (i (- 2 + \sqrt{3}))$
	Commented by infinityaction last updated on 17/Mar/22

	$t h a n k y o u s i r$
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