prove-that-0-pi-2-ln-sinx-dx-pi-2-ln2-

Question Number 75960 by Tony Lin last updated on 21/Dec/19

p r o v e t h a t \int_{0}^{\frac{π}{2}} l n (s i n x) d x = - \frac{π}{2} l n 2

Commented by Kunal12588 last updated on 21/Dec/19

n o w s o l v e \int_{0}^{\frac{π}{2}} l n (c o s x) d x

Commented by Tony Lin last updated on 23/Dec/19

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

l e t θ = \frac{π}{2} - x, d θ = - d x

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

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

Answered by Kunal12588 last updated on 21/Dec/19

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

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

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

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

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

l e t t = 2 x \Rightarrow d x = \frac{1}{2} d t

x \to 0 \Rightarrow t \to 0

x \to \frac{π}{2} \Rightarrow t \to π

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

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

\Rightarrow 2 I = I - \frac{π}{2} l n 2

\Rightarrow I = - \frac{π}{2} l n 2 p r o v e d

Commented by Tony Lin last updated on 23/Dec/19

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