proporsed by m.n july 1790 ∫_0 ^π ln(1−(1/2)sinx)dx


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Question Number 113188 by mathdave last updated on 11/Sep/20

$p r o p o r s e d b y m . n j u l y 1790$ $\int_{0}^{π} \ln (1 - \frac{1}{2} \sin x) d x$
	Answered by mathdave last updated on 11/Sep/20

	$m y s o l u t i o n t o$ $I = \int_{0}^{π} \ln (1 - \frac{1}{2} \sin x) d x = 2 \int_{0}^{\frac{π}{2}} \ln (1 - \frac{1}{2} \sin x) d x . . . . . (1)$ $a n d J = 2 \int_{0}^{\frac{π}{2}} \ln (1 + \frac{1}{2} \sin x) d x . . . . . . . . . . . (2)$ $a d d i n g (1) a n d (2)$ $I + J = 2 \int_{0}^{\frac{π}{2}} \ln (1 - \frac{\sin^{2} x}{4}) d x = 2 \int_{0}^{\frac{π}{2}} \ln (\cos^{2} x + \sin^{2} x - \frac{\sin^{2} x}{4}) d x$ $I + J = 2 \int_{0}^{\frac{π}{2}} \ln (\cos^{2} x + \frac{3}{4} \sin^{2} x) d x$ $r e c a l l t h a t \int_{0}^{\frac{π}{2}} \ln (a^{2} \cos^{2} x + b^{2} \sin^{2} x) d x = π \ln (\frac{a + b}{2})$ $I + J = 2 π \ln (\frac{1}{2} - \frac{\sqrt{3}}{4}) = 2 π \ln (\frac{2 + \sqrt{3}}{4}) . . . . . . (x)$ $s u b t r a c t i n g (1) f r o m (2)$ $J - I = 2 \int_{0}^{\frac{π}{2}} \ln (\frac{1 + \frac{1}{2} \sin x}{1 - \frac{1}{2} \sin x}) d x (a p p l y f a y n m a n n t r i c k)$ $J - I = 2 \int_{0}^{\frac{π}{2}} d x \int_{- \frac{1}{2}}^{\frac{1}{2}} (\frac{\sin x}{1 + y \sin x}) d x$ $J - I = 2 \int_{0}^{\frac{π}{2}} d x [\int_{- \frac{1}{2}}^{0} (\frac{\sin x}{1 + y \sin x}) d y + \int_{0}^{\frac{1}{2}} (\frac{\sin x}{1 + y \sin x}) d y]$ $p u t y = - y$ $J - I = 2 \int_{0}^{\frac{π}{2}} d x [\int_{0}^{\frac{1}{2}} (\frac{\sin x}{1 - y \sin x}) d y + \int_{0}^{\frac{π}{2}} (\frac{\sin x}{1 + y \sin x}) d y]$ $J - I = 2 \int_{0}^{\frac{π}{2}} d x [\int_{0}^{\frac{1}{2}} (\frac{\sin x (1 + y \sin x + 1 - y \sin x)}{(1 - y \sin x) (1 + y \sin x)}) d y]$ $J - I = 4 \int_{0}^{\frac{π}{2}} d x \int_{0}^{\frac{1}{2}} (\frac{\sin x}{1 - y^{2} \sin^{2} x}) d y (b u t \sin^{2} x = 1 - \cos^{2} x)$ $J - I = 4 \int_{0}^{\frac{π}{2}} d x \int_{0}^{\frac{1}{2}} (\frac{\sin x}{1 - y^{2} + y^{2} \cos^{2} x}) d y$ $J - I = 4 \int_{0}^{\frac{1}{2}} d y \int_{0}^{\frac{π}{2}} (\frac{\sin x}{\sqrt{1 - y^{2}} + {(y \cos x)}^{2}}) d x$ $l e t Z = y \cos x, A = \sqrt{1 - y^{2}}, b u t \int \frac{1}{A^{2} + Z^{2}} d Z = \frac{1}{A} \tan^{- 1} (\frac{Z}{A})$ $J - I = - 4 \int_{0}^{\frac{1}{2}} d y [\frac{1}{y \sqrt{1 - y^{2}}} \tan^{- 1} {(\frac{y \cos x}{\sqrt{1 - y^{2}}})}_{0}^{\frac{π}{2}}]$ $J - I = 4 \int_{0}^{\frac{1}{2}} [\frac{\tan^{- 1} (\frac{y}{\sqrt{1 - y^{2}}})}{y \sqrt{1 - y^{2}}}] d y (l e t y = \sin y)$ $J - I = 4 \int_{0}^{\frac{π}{6}} \frac{y}{\sin y} d y (u s i n g I B P) n o t e \int \frac{1}{\sin y} d y = \ln (\tan \frac{y}{2})$ $J - I = 4 {[y \ln (\tan \frac{y}{2})]}_{0}^{\frac{π}{6}} - 4 \int_{0}^{\frac{π}{6}} \ln (\tan \frac{y}{2}) d y$ $J - I = 4 ∙ \frac{π}{6} \ln (\tan \frac{π}{12}) - 8 \int_{0}^{\frac{π}{12}} \ln (\tan y) d y$ $n o t e s \tan \frac{A}{2} = \sqrt{\frac{1 - \cos 2 A}{1 + \cos 2 A}} a n d f r o m$ $l e m m a (2) \int_{0}^{\frac{π}{12}} \ln (\tan y) d y = - \frac{2}{3} G$ $∵ J - I = \frac{π}{3} \ln [\frac{1 - \cos \frac{π}{6}}{1 + \cos \frac{π}{6}}] - 8 [- \frac{2}{3} G]$ $J - I = \frac{π}{3} [\frac{2 - \sqrt{3}}{2 + \sqrt{3}}] + \frac{16}{3} G = \frac{16}{3} - \frac{2 π}{3} \ln (2 + \sqrt{3}) . . . . . (x x)$ $∵ I = \int_{0}^{π} \ln (1 - \frac{1}{2} \sin x) d x = \frac{J + I - (J - I)}{2}$ $I = \frac{2 π \ln (\frac{2 + \sqrt{3}}{4}) - \frac{16}{3} G + \frac{2 π}{3} \ln (2 + \sqrt{3})}{2}$ $I = π \ln (\frac{2 + \sqrt{3}}{4}) - \frac{8}{3} G + \frac{π}{3} \ln (2 + \sqrt{3})$ $∵ \int_{0}^{π} (1 - \frac{1}{2} \sin x) d x = \frac{4 π}{3} \ln (2 + \sqrt{3}) - \frac{8}{3} G - 2 π \ln 2$ $b y m a t h d a v e (12 / 09 / 2020)$
	Commented by mnjuly1970 last updated on 11/Sep/20

	$e x e l l e n t . v e r y n i c e$ $g r a t e f u l m r d a v e . .$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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