0-pi-2-log-sin-x-dx-

Question Number 89161 by M±th+et£s last updated on 15/Apr/20

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

Commented by niroj last updated on 15/Apr/20

I = \int_{0}^{\frac{π}{2}} \log \sin x dx \dots . . (i)

= \int_{0}^{\frac{π}{2}} \log \sin (\frac{π}{2} - x) dx [∵ \int_{0}^{\boldsymbol a} \boldsymbol f (\boldsymbol c) \boldsymbol dx = \int_{0}^{\boldsymbol a} \boldsymbol f (\boldsymbol a - \boldsymbol c) \boldsymbol dx]

I = \int_{0}^{\frac{π}{2}} \log \cos x dx \dots . . (ii)

added (i) + (ii)

2 I = \int_{0}^{\frac{π}{2}} (\log \sin x + \log \cos x) dx

2 I = \int_{0}^{\frac{π}{2}} \log \sin x . \cos xdx

2 I = \int_{0}^{\frac{π}{2}} \log (\frac{2 \sin xcos x}{2}) dx

2 I = {\int_{0}}^{\frac{π}{2}} \log \sin 2 x dx - \int_{0}^{\frac{π}{2}} \log 2 dx

put 2 x = t

2 dx = dt

dx = \frac{dt}{2}

if x = \frac{π}{2} then t = π

if x = 0 then t = 0

\int_{0}^{π} \log \sin t . \frac{dt}{2}

= \frac{1}{2} \int_{0}^{π} \log sint dt

= \frac{1}{2} \times 2 \int_{0}^{\frac{π}{2}} \log sint dt

∴ \int_{0}^{\frac{π}{2}} \log sint dt = \int_{0}^{\frac{π}{2}} \log \sin x dx = I

2 I = I - \log 2 \int_{0}^{\frac{π}{2}} dx

2 I - I = - \log 2 {[x]}_{0}^{\frac{π}{2}}

I = \log 2^{- 1} [\frac{π}{2} - 0]

I = \frac{π}{2} \log \frac{1}{2} / / .

Commented by M±th+et£s last updated on 15/Apr/20

t h a n x f o r t h e s o l u t i o n s

Answered by TANMAY PANACEA. last updated on 15/Apr/20

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

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

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

\boldsymbol n o w \boldsymbol m a i n \boldsymbol p o i n t \dots

\boldsymbol t = 2 \boldsymbol x

\int_{0}^{π} l o g s i n t \times \frac{d t}{2}

\frac{1}{2} \int_{0}^{π} l o g s i n t d t

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

u s i n g \int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x

w h e n f (2 a - x) = f (x)

h e r e a = \frac{π}{2}

s o s i n (2 \times \frac{π}{2} - x) = s i n x

\boldsymbol n o w \dots

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

2 I = I - \frac{π}{2} l o g 2

I = - \frac{π}{2} l o g 2