find A(x)=∫_0 ^(π/2) ln(1−xsin^2 θ)dθ with ∣x∣<1


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Question Number 72397 by mathmax by abdo last updated on 28/Oct/19

$f i n d A (x) = \int_{0}^{\frac{π}{2}} l n (1 - {x s i n}^{2} θ) d θ w i t h ∣ x ∣< 1$
Commented bymathmax by abdo last updated on 28/Oct/19

$a t f o r m o f s e r i e w e h a v e {l n}^{^{'}} (1 - u) = \frac{- 1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n} \Rightarrow$ $l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1} + c (c = 0)$ $= - \sum_{n = 1}^{\infty} \frac{u^{n}}{n} \Rightarrow l n (1 - {x s i n}^{2} θ) = - \sum_{n = 1}^{\infty} \frac{x^{n} {s i n}^{2 n} θ}{n} \Rightarrow$ $A (x) = - \sum_{n = 1}^{\infty} (\int_{0}^{\frac{π}{2}} {s i n}^{2 n} θ d θ) \frac{x^{n}}{n} = - \sum_{n = 1}^{\infty} \frac{W_{n}}{n} x^{n}$ $w i t h W_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{2 n} d θ (w a l l i s i n t e g r a l)$
Commented bymathmax by abdo last updated on 28/Oct/19
$let try the parametric method we have A(x)=∫_0 ^(π/2) ln(1−x((1−cos(2θ))/2))dθ =∫_0 ^(π/2) ln(2−x+xcosθ)dθ−(π/2)ln(2) let f(x)=∫_0 ^(π/2) ln(2−x+xcos(2θ))dθ ⇒f^′ (x)=∫_0 ^(π/2) ((−1+cos(2θ))/(2−x +xcos(2θ)))dθ =_(2θ=t) (1/2)∫_0 ^π ((−1+cost)/(2−x +xcost))dt =_(tan((t/2))=u) (1/2)∫_0 ^∞ ((((1−u^2 )/(1+u^2 ))−1)/(2−x+x((1−u^2 )/(1+u^2 ))))((2du)/(1+u^2 )) =(1/2)∫_0 ^∞ ((−4u)/((1+u^2 )^2 {2−x+x((1−u^2 )/(1+u^2 ))}))du =−2∫_0 ^∞ (u/((2−x)(1+u^2 )^2 +x(1−u^4 )))du =−2∫_0 ^∞ (u/((2−x)(u^4 +2u^2 +1)+x−xu^4 )) =−2∫_0 ^∞ ((udu)/((2−2x)u^4 +(4−2x)u^2 +2)) =−∫_0 ^∞ ((udu)/((1−x)u^4 +(2−x)u^2 +1)) =∫_0 ^∞ ((udu)/((x−1)u^4 +(x−2)u^2 −1)) let decompose F(u) =(u/((x−1)u^4 +(x−2)u^2 −1)) (x−1)u^4 +(x−2)u^2 −1 =0 ⇒(x−1)z^2 +(x−2)z −1=0 with z =u^2 Δ=(x−2)^2 −4(x−1)(−1) =x^2 −4x +4+4x−4=x^2 ⇒ z_1 =((−x+2+∣x∣)/(2(x−1))) and z_2 =((−x+2−∣x∣)/(2(x−1))) case1 x≥0 and x≠1 ⇒z_1 =(1/(x−1)) and z_2 =((−2x+2)/(2(x−1))) =−1 ⇒ F(u) =(u/((x−1(z−z_1 )(z−z_2 ))) =(u/((x−1)(u^2 −(1/(x−1)))(u^2 +1))) if x>1 ⇒F(u) =(u/((x−1)(u−(1/(√(x−1))))(u+(1/(√(x−1))))(u^2 +1))) =(a/(u−(1/(√(x−1))))) +(b/(u+(1/(√(x−1))))) +((cu +d)/(u^2 +1)) ...be continued...$
$l e t t r y t h e p a r a m e t r i c m e t h o d w e h a v e$ $A (x) = \int_{0}^{\frac{π}{2}} l n (1 - x \frac{1 - c o s (2 θ)}{2}) d θ = \int_{0}^{\frac{π}{2}} l n (2 - x + x c o s θ) d θ - \frac{π}{2} l n (2)$ $l e t f (x) = \int_{0}^{\frac{π}{2}} l n (2 - x + x c o s (2 θ)) d θ \Rightarrow f^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{- 1 + c o s (2 θ)}{2 - x + x c o s (2 θ)} d θ$ $=_{2 θ = t} \frac{1}{2} \int_{0}^{π} \frac{- 1 + c o s t}{2 - x + x c o s t} d t =_{t a n (\frac{t}{2}) = u} \frac{1}{2} \int_{0}^{\infty} \frac{\frac{1 - u^{2}}{1 + u^{2}} - 1}{2 - x + x \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \frac{1}{2} \int_{0}^{\infty} \frac{- 4 u}{{(1 + u^{2})}^{2} {2 - x + x \frac{1 - u^{2}}{1 + u^{2}}}} d u$ $= - 2 \int_{0}^{\infty} \frac{u}{(2 - x) {(1 + u^{2})}^{2} + x (1 - u^{4})} d u$ $= - 2 \int_{0}^{\infty} \frac{u}{(2 - x) (u^{4} + 2 u^{2} + 1) + x - {x u}^{4}}$ $= - 2 \int_{0}^{\infty} \frac{u d u}{(2 - 2 x) u^{4} + (4 - 2 x) u^{2} + 2}$ $= - \int_{0}^{\infty} \frac{u d u}{(1 - x) u^{4} + (2 - x) u^{2} + 1} = \int_{0}^{\infty} \frac{u d u}{(x - 1) u^{4} + (x - 2) u^{2} - 1}$ $l e t d e c o m p o s e F (u) = \frac{u}{(x - 1) u^{4} + (x - 2) u^{2} - 1}$ $(x - 1) u^{4} + (x - 2) u^{2} - 1 = 0 \Rightarrow (x - 1) z^{2} + (x - 2) z - 1 = 0$ $w i t h z = u^{2}$ $Δ = {(x - 2)}^{2} - 4 (x - 1) (- 1) = x^{2} - 4 x + 4 + 4 x - 4 = x^{2} \Rightarrow$ $z_{1} = \frac{- x + 2 + ∣ x ∣}{2 (x - 1)} a n d z_{2} = \frac{- x + 2 - ∣ x ∣}{2 (x - 1)}$ $c a s e 1 x ⩾ 0 a n d x \neq 1 \Rightarrow z_{1} = \frac{1}{x - 1} a n d z_{2} = \frac{- 2 x + 2}{2 (x - 1)} = - 1 \Rightarrow$ $F (u) = \frac{u}{(x - 1 (z - z_{1}) (z - z_{2})} = \frac{u}{(x - 1) (u^{2} - \frac{1}{x - 1}) (u^{2} + 1)}$ $i f x > 1 \Rightarrow F (u) = \frac{u}{(x - 1) (u - \frac{1}{\sqrt{x - 1}}) (u + \frac{1}{\sqrt{x - 1}}) (u^{2} + 1)}$ $= \frac{a}{u - \frac{1}{\sqrt{x - 1}}} + \frac{b}{u + \frac{1}{\sqrt{x - 1}}} + \frac{c u + d}{u^{2} + 1} . . . b e c o n t i n u e d . . .$
Commented bymind is power last updated on 29/Oct/19

$this will worck sir since A (0) = 0, nice worck$
Commented bymathmax by abdo last updated on 29/Oct/19

$w e h a v e A (0) = 0$
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