0-pi-2-ln-cos-x-dx-

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

\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)