𝛀 =∫_( 0) ^( (𝛑/4)) x log (1 + tanx) dx = ?


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Question Number 156923 by MathSh last updated on 17/Oct/21

$Ω = \underset{0}{\int^{\frac{π}{4}}} x \log (1 + \tan x) dx = ?$
	Answered by mindispower last updated on 18/Oct/21
	$∫_0 ^(π/4) xln(cos(x))dx=a...? ln(cos(x))=−Σ_(k≥1) (((−1)^k cos(2kx))/k)−ln(2)..fourie sere a=−Σ_(k≥1) (((−1)^k )/k)∫_0 ^(π/4) xcos(2kx)dx−ln(2)∫_0 ^(π/4) xdx =−Σ_(k≥1) (((−1)^k )/k){[((xsin(2kx))/(2k))]_0 ^(π/4) +(1/(4k^2 ))[cos(2kx)]_0 ^(π/4) −(π^2 /(32))ln(2) =−(Σ_(k≥1) (((−1)^k sin(((kπ)/2)))/(2k^2 ))+(((−1)^k cos(((kπ)/2)))/(4k^3 ))) spit k=2m,k=2m+1we get withe sin(πm)=0 sin(πm+(π/2))=(−1)^m ,cos(mπ)=(−1)^m ,cos(mπ+(π/2))=(−1)^m =Σ_(m≥0) (((−1)^m )/(2(2m+1)^2 ))−Σ_(m≥1) (((−1)^m )/(32m^3 )) =(1/2)𝛃(−1)−(1/(32))Li_3 (−1)=(G/2)+(3/(4.32))ζ(3)−((π^2 ln(2))/(32)) ∫_0 ^(π/4) ln(cos(x))dx,∫_0 ^(π/4) ln(tg(x))=G ∫_0 ^(π/2) ln(sinx)dx=−(π/2)ln(2) cos(x)=(1/2)ln(((sin(x)cos(x))/(sin(x)))cos(x)),x∈[0,(π/2)] =(1/2)ln(sin(2x))−((ln(tg(x)))/2) ⇒∫_0 ^(π/4) ln(cos(x))=(π/8)ln(2)−(G/2) Ω=∫_0 ^(π/4) xln((((√2)sin(x+(π/4)))/(cos(x))))dx =((ln(2))/2)∫_0 ^(π/4) xdx+∫_0 ^(π/4) xln(sin(x+(π/4)))dx−∫_0 ^(π/4) xln(cos(x))dx =((ln(2))/(32))π^2 +∫_0 ^(π/4) (π/4)ln(cos(x))dx−2∫_0 ^(π/4) xln(cos(x))dx =((ln(2))/(32))π^2 +(π/4)(((πln2)/8)−(G/2))−2((G/2)+((3ζ(1))/(128))−((π^2 ln(2))/(32))) =−((π/8)+1)G+((π^2 ln(2))/8)−(3/(64))ζ(3)$
	$\int_{0}^{\frac{π}{4}} x l n (c o s (x)) d x = a . . . ?$ $l n (c o s (x)) = - \sum_{k ⩾ 1} \frac{{(- 1)}^{k} c o s (2 k x)}{k} - l n (2) . . f o u r i e s e r e$ $a = - \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} \int_{0}^{\frac{π}{4}} x c o s (2 k x) d x - l n (2) \int_{0}^{\frac{π}{4}} x d x$ $= - \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} {{[\frac{x s i n (2 k x)}{2 k}]}_{0}^{\frac{π}{4}} + \frac{1}{4 k^{2}} {[c o s (2 k x)]}_{0}^{\frac{π}{4}} - \frac{π^{2}}{32} l n (2)$ $= - (\sum_{k ⩾ 1} \frac{{(- 1)}^{k} s i n (\frac{k π}{2})}{2 k^{2}} + \frac{{(- 1)}^{k} c o s (\frac{k π}{2})}{4 k^{3}})$ $s p i t k = 2 m, k = 2 m + 1 w e g e t w i t h e s i n (π m) = 0$ $s i n (π m + \frac{π}{2}) = {(- 1)}^{m}, c o s (m π) = {(- 1)}^{m}, c o s (m π + \frac{π}{2}) = {(- 1)}^{m}$ $= \sum_{m ⩾ 0} \frac{{(- 1)}^{m}}{2 {(2 m + 1)}^{2}} - \sum_{m ⩾ 1} \frac{{(- 1)}^{m}}{32 m^{3}}$ $= \frac{1}{2} β (- 1) - \frac{1}{32} {L i}_{3} (- 1) = \frac{G}{2} + \frac{3}{4.32} ζ (3) - \frac{π^{2} l n (2)}{32}$ $\int_{0}^{\frac{π}{4}} l n (c o s (x)) d x, \int_{0}^{\frac{π}{4}} l n (t g (x)) = G$ $\int_{0}^{\frac{π}{2}} l n (s i n x) d x = - \frac{π}{2} l n (2)$ $c o s (x) = \frac{1}{2} l n (\frac{s i n (x) c o s (x)}{s i n (x)} c o s (x)), x \in [0, \frac{π}{2}]$ $= \frac{1}{2} l n (s i n (2 x)) - \frac{l n (t g (x))}{2}$ $\Rightarrow \int_{0}^{\frac{π}{4}} l n (c o s (x)) = \frac{π}{8} l n (2) - \frac{G}{2}$ $Ω = \int_{0}^{\frac{π}{4}} x l n (\frac{\sqrt{2} s i n (x + \frac{π}{4})}{c o s (x)}) d x$ $= \frac{l n (2)}{2} \int_{0}^{\frac{π}{4}} x d x + \int_{0}^{\frac{π}{4}} x l n (s i n (x + \frac{π}{4})) d x - \int_{0}^{\frac{π}{4}} x l n (c o s (x)) d x$ $= \frac{l n (2)}{32} π^{2} + \int_{0}^{\frac{π}{4}} \frac{π}{4} l n (c o s (x)) d x - 2 \int_{0}^{\frac{π}{4}} x l n (c o s (x)) d x$ $= \frac{l n (2)}{32} π^{2} + \frac{π}{4} (\frac{π l n 2}{8} - \frac{G}{2}) - 2 (\frac{G}{2} + \frac{3 ζ (1)}{128} - \frac{π^{2} l n (2)}{32})$ $= - (\frac{π}{8} + 1) G + \frac{π^{2} l n (2)}{8} - \frac{3}{64} ζ (3)$
	Commented by MathSh last updated on 18/Oct/21

	$Perfect dear Ser, thank you so much$
	Commented by mindispower last updated on 19/Oct/21

	$w i t h e P l e a s u r$
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