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

Question Number 52946 by gunawan last updated on 15/Jan/19

\underset{0}{\int^{π / 2}} \log \sin 2 x d x =

Commented by maxmathsup by imad last updated on 15/Jan/19

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

\Rightarrow 2 A = \int_{0}^{\frac{π}{2}} l o g (s i n t) d t + \int_{\frac{π}{2}}^{π} l o g (s i n t) d t b u t w e h a v e p r o v e d t h a t

\int_{0}^{\frac{π}{2}} l o g (s i n t) d t = - \frac{π}{2} l n (2) a l s o \int_{\frac{π}{2}}^{π} l o g (s i n t) d t =_{t = \frac{π}{2} + u} \int_{0}^{\frac{π}{2}} l o g (s i n (\frac{π}{2} + u)) d u

= \int_{0}^{\frac{π}{2}} l o g (c o s u) d u = - \frac{π}{2} l n (2) (r e s u l t p r o v e d) \Rightarrow

2 A = - π l n (2) \Rightarrow A = - \frac{π}{2} l n (2) .

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

t = 2 x

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

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

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

[n o w \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)]

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

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

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

2 I = \int_{0}^{\frac{π}{2}} l n s i n t + l n c o s t d t \frac{}{} p

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

= \int_{0}^{\frac{π}{2}} [l n s i n 2 t - l n 2] d t

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

2 I = I - l n 2 \times ∣ t ∣_{0}^{\frac{π}{2}}

I = - l n 2 \times \frac{π}{2} p l s c h e c k \dots

Commented by gunawan last updated on 16/Jan/19

Nice Sir

thank you so much

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

m o s t w e l c o m e \dots