prove-0-2-log-sinx-dx-pi-2-log-1-2-

Question Number 179852 by arup last updated on 03/Nov/22

p r o v e \int_{0}^{\frac{π}{2}} l o g (s i n x) d x = \frac{π}{2} l o g \frac{1}{2}

Commented by som(math1967) last updated on 03/Nov/22

I = \int_{0}^{\frac{π}{2}} l o g [s i n (\frac{π}{2} - x)] d x

2 I = \int_{0}^{\frac{π}{2}} l o g (s i n x) + l o g (c o s x) d x

= \int_{0}^{\frac{π}{2}} l o g (\frac{2 s i n x c o s x}{2}) d x

2 I = \int_{0}^{\frac{π}{2}} l o g (s i n 2 x) d x - \int_{0}^{\frac{π}{2}} l o g 2 d x

2 I = \int_{0}^{\frac{π}{2}} l o g (s i n 2 x) d x - \frac{π}{2} l o g 2

l e t 2 x = t

2 I = \frac{1}{2} \int_{0}^{π} l o g (s i n t) d t - \frac{π}{2} l o g 2

2 I = \frac{1}{2} \times 2 \int_{0}^{\frac{π}{2}} l o g (s i n t) d t - \frac{π}{2} l o g 2

[\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x

i f (2 a - x) = f (x)]

2 I = \int_{0}^{\frac{π}{2}} l o g (s i n t) d t + \frac{π}{2} l o g \frac{1}{2}

2 I = \int_{0}^{\frac{π}{2}} l o g (s i n x) d x + \frac{π}{2} l o g \frac{1}{2}

[\int_{a}^{b} f (t) d t = \int_{a}^{b} f (x) d x]

2 I - I = \frac{π}{2} l o g \frac{1}{2}

∴ I = \frac{π}{2} {l o g}_{e} \frac{1}{2}

Commented by Frix last updated on 03/Nov/22

{can}^{'} t prove it but {it}^{'} s the same as

\underset{0}{\int^{\frac{π}{2}}} \frac{x}{\tan x} d x

Commented by som(math1967) last updated on 03/Nov/22

a m i c o r r e c t ?

Commented by CElcedricjunior last updated on 03/Nov/22

\int_{0}^{\frac{π}{2}} l o g (s i n x) dx = \frac{π}{2} l o g \frac{1}{2}

k = \int_{0}^{π / 2} \log (sinx) dx

o n a e s i n x = c o s (\frac{π}{2} - x)

k = \int_{0}^{π / 2} l o g (c o s (- x + \frac{π}{2})) d x

p o s o n s - x + \frac{π}{2} = t \Leftrightarrow d x = - d t

q d : {\begin{cases} x - > 0 \\ x - > π / 2 \end{cases} => {\begin{cases} t - > π / 2 \\ t - > 0 \end{cases}

k = \int_{0}^{π / 2} l o g (c o s t) d t = l

=> k - l = 0

k + l = \int_{0}^{π / 2} l o g (s i n x c o s x) d x

= \int_{0}^{π / 2} [l o g (\frac{1}{2}) + l o g (s i n 2 x)] d x

= \int_{0}^{π / 2} l o g (s i n 2 x) d x + {[x]}_{0}^{\frac{π}{2}} l o g (\frac{1}{2})

k + l = \frac{π}{2} l o g (\frac{1}{2}) + \int_{0}^{π / 2} l o g (s i n 2 x) d x

k + l = \frac{π}{2} l o g (\frac{1}{2}) + \int_{0}^{π / 2} l o g (c o s (2 x - \frac{π}{2})) d x

p o s o n s 2 x - \frac{π}{2} = t => d x = \frac{1}{2} d t

q d : {\begin{cases} x - > 0 \\ x - > π / 2 \end{cases} => {\begin{cases} t - > - \frac{π}{2} \\ t - > \frac{π}{2} \end{cases}

k + l = \frac{π}{2} \log (\frac{1}{2}) + \frac{1}{2} \int_{- \frac{π}{2}}^{π / 2} l o g (c o s t) d t

k + l = \frac{π}{2} l o g (1 / 2) + \int_{0}^{π / 2} l o g (c o s t) d t

k + l = \frac{π}{2} l o g (1 / 2) + l

=> k = \frac{π}{2} l o g (\frac{1}{2})

\int_{0}^{\frac{π}{2}} l o g (s i n x) d x = \frac{π}{2} l o g (\frac{1}{2})

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