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Question Number 25821 by abdo imad last updated on 15/Dec/17

a n s w e r t o q 25796 f_{} i n d t h e v a l u e o f f (x) = \int_{0^{}}^{π_{}} l n (1 + x c o s θ) d θ w i t h 0 < x < 1 \partial f / \partial x = \int_{0}^{π} c o s θ {(1 + x c o s θ)}^{- 1} d θ = π x^{- 1} - x^{- 1} \int_{0}^{π} {(1 + x c o s θ)}^{- 1} d θ a n d b y t h e c h a n g e e n t t a n θ = u t h e n t h e c h a n g e m e n t u = {((1 + x) {(1 + x)}^{- 1^{}})}^{} . t . w e f i n d \int_{0}^{θ} {(1 + x c o s θ)}^{- 1} d θ = π {(1 - x^{2_{}})}^{- 1 / 2}

s o \partial f / \partial x = π x^{- 1} - π x^{- 1} (1 - x^{- 1 / 2}) s o f (x) = π l n (x) - π \int^{x} t^{- 1} {(1 - t^{2})}^{- 1 / 2} d t

b u t \int^{x} t^{- 1} (1 - t^{2}) d t = - 1 . 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2} . {(1 - {(1 - x^{2})}^{1 / 2})}^{- 1})

a n d f (x) = π l n (x) + π . 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2} ({(1 - {(1 - x^{2})}^{1 / 2})}^{- 1} + β

β = f (1) = \int_{0}^{π =} l n (1 + c o s θ) d θ = - π l n (2) s o

\int_{0}^{π} (1 + x c o s θ) d θ = π l n x + π 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2^{}}) ({(1 - {(1 - x^{2})}^{1 / 2})}^{- 1} - π l n (2) .