∫ln (sin (x))dx=?


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Question Number 114880 by manuel2456 last updated on 21/Sep/20

$\int \ln (\sin (x)) d x = ?$
Commented by MJS_new last updated on 21/Sep/20

$similar to question 113634$
	Answered by mathdave last updated on 23/Sep/20

	$s o l u t i o n$ $f r o m b i n o m i a l t h e o r e m$ $l e t \frac{1}{1 - y} = \underset{n = 1}{\sum^{\infty}} y^{n - 1} o r \ln (1 - y) = - \underset{n = 1}{\sum^{\infty}} \frac{y^{n}}{n}$ $p u t y = e^{i 2 x}$ $\ln (1 - e^{i 2 x}) = - \underset{n = 1}{\sum^{\infty}} \frac{e^{i 2 n x}}{n}$ $\ln (1 - (\cos 2 x + i \sin 2 x)) = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} (\cos (2 n x) + i \sin (2 n x))$ $b u t {2 \sin}^{2} x = 1 - \cos 2 x a n d \sin 2 x = 2 \sin x \cos x$ $\ln ({2 \sin}^{2} x - i 2 \sin x \cos x) = - \underset{n = 1}{\sum^{\infty}} \frac{\cos (2 n x)}{n} - i \underset{n = 1}{\sum^{\infty}} \frac{\sin (2 n x)}{n}$ $\ln (2 \sin x (\sin x - i \cos x)) = - \underset{n = 1}{\sum^{\infty}} \frac{\cos (2 n x)}{n} - i \underset{n = 1}{\sum^{\infty}} \frac{\sin (2 n x)}{n}$ $\ln (2 \sin x) + \ln (\sin x - i \cos x) = - \underset{n = 1}{\sum^{\infty}} \frac{\cos (2 n x)}{n} - i \underset{n = 1}{\sum^{\infty}} \frac{\sin (2 n x)}{n}$ $l e t$ $\ln (- i (- \frac{1}{i} \sin x + \cos x)) = \ln (- i (\cos x - \frac{1}{i} \sin x)) = \ln (- i (\cos x + i \sin x)) = \ln (- i . e^{i x})$ $b u t$ $\ln (- i . e^{i x}) = \ln (- 1) + \ln (e^{i x}) = i x + \ln (- i) = i x - i \frac{π}{2} = - i (\frac{π}{2} - x)$ $∴ \ln (2 \sin x) - i (\frac{π}{2} - x) = - \underset{n = 1}{\sum^{\infty}} \frac{\cos (2 n x)}{n} - i \underset{n = 1}{\sum^{\infty}} \frac{\sin (2 n x)}{n}$ $b y e q u a t i n g r e a l a n d i m a g i n a r y p a r t$ $\ln (2 \sin x) = - \underset{n = 1}{\sum^{\infty}} \frac{\cos (2 n x)}{n}$ $\ln 2 + \ln (\sin x) = - \underset{n = 1}{\sum^{\infty}} \frac{\cos (2 n x)}{n}$ $\ln (\sin x) = - \ln 2 - \underset{n = 1}{\sum^{\infty}} \frac{\cos (2 n x)}{n} . . . . . . (1)$ $∵ \int \ln (\sin x) d x = - \ln 2 \int d x - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int \cos (2 n x) d x$ $∵ \int \ln (\sin x) d x = - x \ln 2 - \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n^{2}} \sin (2 n x)$ $b y m a t h d a v e (23 / 09 / 2020)$
	Commented by Tawa11 last updated on 06/Sep/21

	$grest sir$
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