find-G-1-4-0-pi-2-ln-1-sin-x-1-sin-x-dx-

Question Number 212319 by Ghisom last updated on 09/Oct/24

find

G = \frac{1}{4} \underset{0}{\int^{π / 2}} \ln \frac{1 + \sin x}{1 - \sin x} d x

Commented by Spillover last updated on 10/Oct/24

\frac{π}{4} \ln 2 r i g h t ?

Commented by Frix last updated on 10/Oct/24

\frac{π}{4} \ln 2 \approx . 544

G \approx . 916

Answered by Spillover last updated on 10/Oct/24

= \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\frac{1 + \sin x}{1 - \sin x})

= \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\frac{1 + \sin x}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x}) = \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln \frac{{(1 + \sin x)}^{2}}{\cos^{2} x}

\frac{1}{4} \int_{0}^{\frac{π}{2}} \ln \frac{{(1 + \sin x)}^{2}}{\cos^{2} x} = \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln {(1 + \sin x)}^{2} - \ln \cos^{2} x

\frac{1}{4} \int_{0}^{\frac{π}{2}} \ln {(1 + \sin x)}^{2} - \ln \cos^{2} x

\frac{1}{4} \int_{0}^{\frac{π}{2}} [2 \ln (1 + \sin x) - 2 \ln \cos x] d x

[\frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (1 + \sin x) d x] - [\frac{1}{2} \int_{0}^{\frac{π}{2}} \ln \cos x] d x

[\frac{1}{2} \int_{0}^{\frac{π}{2}} \ln \cos x] d x = - \frac{π}{2} \ln 2

[\frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (1 + \sin x) d x]

\dots .

Answered by Ar Brandon last updated on 11/Oct/24

Ω = \frac{1}{4} \int_{0}^{\frac{π}{2}} \ln (\frac{1 + \sin x}{1 - \sin x}) d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}) d x

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\frac{\sin (\frac{x}{2} + \frac{π}{4})}{\cos (\frac{x}{2} + \frac{π}{4})}) d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \ln (\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}) d x

= \int_{0}^{\frac{π}{4}} \ln (\frac{\cos x}{\sin x}) d x = - \int_{0}^{\frac{π}{4}} \ln (\tan x) d x = G

G = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}}, {Catalan}^{'} s constant .

Commented by Ghisom last updated on 12/Oct/24

yes

Commented by Ar Brandon last updated on 12/Oct/24

Yep, Sir MJS ��

Commented by Spillover last updated on 12/Oct/24

H o w d i d y o u k n o w i f s i r M J s . u s e r n a m e

i s d i f f e r e n t

Commented by Ar Brandon last updated on 12/Oct/24

I can tell from his writings.

Commented by Frix last updated on 12/Oct/24

I^{'} m confused . I thought it was me ? So if {it}^{'} s

not me, then who am I ?

(I^{'} m n o t Ghisom, {Scout}^{'} s honor!)

Commented by Ghisom last updated on 13/Oct/24

if you were me I would see you in my

mirror and v / v . but I see me, at least I

see the same guy {who}^{'} s on the photographs

with my wife .

Commented by Ar Brandon last updated on 13/Oct/24

You're the long-white-bearded old man. Haha!