L-0-4-pi-ln-cosx-dx-

Question Number 208280 by Shrodinger last updated on 10/Jun/24

L = \int_{0}^{\frac{4}{π}} l n (c o s x) d x

Answered by Berbere last updated on 10/Jun/24

= \int_{0}^{\frac{π}{4}} \frac{l n (\frac{c o s (x)}{s i n (x)} . s i n (x) c o s (x)) d x}{2} = \frac{1}{2} \int_{0}^{\frac{π}{4}} l n (c o t (x)) d x + \frac{1}{2} \int_{0}^{\frac{π}{4}} l n (\frac{s i n (2 x)}{2}) d x

t a n (x) \to t; 2 x \to t

= - \frac{1}{2} \int_{0}^{1} \frac{l n (t)}{1 + t^{2}} d t + \frac{1}{4} \int_{0}^{\frac{π}{2}} l n (s i n (y)) d y - \frac{l n (2)}{2} . \frac{π}{4}

= - \frac{1}{2} \int_{0}^{1} \sum_{n ⩾ 0} {(- 1)}^{n} l n (t) t^{2 n} d t + \frac{1}{4} . - \frac{π}{2} l n (2) - \frac{π l n (2)}{8}

= - \frac{1}{2} \sum_{n ⩾ 0} - \frac{1}{{(2 n + 1)}^{2}} - \frac{π l n (2)}{4}; G = \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} = β (- 1) C a t a l a n e c o n s t a n t

= \frac{1}{2} G - \frac{π l n (2)}{4}

Commented by Shrodinger last updated on 11/Jun/24

t h a n k s s i r . .

Commented by Shrodinger last updated on 11/Jun/24

S i r I t i s \int_{0}^{\frac{4}{π}} l n (c o s x) d x