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

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

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

\boldsymbol k = \int_{0}^{\boldsymbol π / 2} \boldsymbol \log (\boldsymbol sinx) \boldsymbol dx

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

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

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

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

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

=> \boldsymbol k - \boldsymbol l = 0

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

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

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

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

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

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

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

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

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

\boldsymbol k + \boldsymbol l = \frac{\boldsymbol π}{2} \boldsymbol l o g (1 / 2) + \boldsymbol l

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

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

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