∫_0 ^(π/2) ln (cos x) dx


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Question Number 105412 by john santu last updated on 28/Jul/20

$\underset{0}{\int^{π / 2}} \ln (\cos x) d x$
	Answered by Dwaipayan Shikari last updated on 28/Jul/20
	$∫_0 ^(π/2) log(cosx)dx=I=∫_0 ^(π/2) log(sinx)dx 2I=∫_0 ^(π/2) log(sinxcosx)dx 2I=∫_0 ^(π/4) log(sin2x)dx−∫_0 ^(π/2) log(2) { ∫_0 ^(π/4) log(sin2x)=∫_0 ^(π/2) log(sint)} 2I=I−∫_0 ^(π/2) log(2) I=−(π/2)log(2)$
	$\int_{0}^{\frac{π}{2}} l o g (c o s x) d x = I = \int_{0}^{\frac{π}{2}} l o g (s i n x) d x$ $2 I = \int_{0}^{\frac{π}{2}} l o g (s i n x c o s x) d x$ $2 I = \int_{0}^{\frac{π}{4}} l o g (s i n 2 x) d x - \int_{0}^{\frac{π}{2}} l o g (2) {\int_{0}^{\frac{π}{4}} l o g (s i n 2 x) = \int_{0}^{\frac{π}{2}} l o g (s i n t)}$ $2 I = I - \int_{0}^{\frac{π}{2}} l o g (2)$ $I = - \frac{π}{2} l o g (2)$
	Commented by bemath last updated on 28/Jul/20

	$\underset{0}{\int^{π / 2}} \log (2) d x ?$
	Commented by Dwaipayan Shikari last updated on 28/Jul/20

	${[x l o g (2)]}_{0}^{\frac{π}{2}} = (\frac{π}{2} - 0) l o g 2 = \frac{π}{2} l o g 2$
	Commented by Dwaipayan Shikari last updated on 28/Jul/20

	$\int_{0}^{\frac{π}{2}} l o g (c o s x) d x$ $= \int_{0}^{\frac{π}{2}} l o g (\frac{e^{i x} + e^{- i x}}{2}) d x$ $= \int_{0}^{\frac{π}{2}} l o g e^{- i x} + l o g (e^{2 i x} + 1) - \int_{0}^{\frac{π}{2}} l o g 2$ $= \int_{0}^{\frac{π}{2}} - i x + \int l o g (e^{2 i x} + 1) - \frac{π}{2} l o g (2)$ $= - \frac{i π^{2}}{8} - \frac{π}{2} l o g (2) + \int_{0}^{\frac{π}{2}} l o g (e^{2 i x} + 1)$ $= \frac{- i π^{2}}{8} - \frac{π}{2} l o g (2) + \frac{i π^{2}}{8} = - \frac{π}{2} l o g (2)$
	Commented by Dwaipayan Shikari last updated on 28/Jul/20

	Commented by Dwaipayan Shikari last updated on 28/Jul/20

	$F r o m w o l f r a m a l p h a$
	Answered by bemath last updated on 29/Jul/20

	Commented by mathmax by abdo last updated on 29/Jul/20

	$no sir the correct answer is \int_{0}^{\frac{π}{2}} \ln (sinu) du = - \frac{π}{2} \ln (2)$
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