answer to q25796 f_ ind the value off(x)= ∫_0^ ^π_ ln(1+xcosθ)dθ with 0<x<1 ∂f/∂x= ∫_0 ^π cosθ(1+xcosθ)^(−1) dθ=πx^(−1) −x^(−1) ∫_0 ^π (1+xcosθ)^(−1) dθand by the changeent tanθ=u then the changement u=((1+x)(1+x)^(−1^ ) )^ .t.we find ∫_0 ^θ (1+xcosθ)^(−1) dθ =π(1−x^2_ )^(−1/2) so ∂f/∂x=πx^(−1) −πx^(−1) (1−x^(−1/2) ) so f(x)= π ln(x)−π∫^x t^(−1) (1−t^2 )^(−1/2) dt but ∫^x t^(−1) (1−t^2 )dt=−1. 2^(−1) ln((1+(1−x^2 )^(1/2) .(1−(1−x^2 )^(1/2) )^(−1) ) and f(x)=πln(x)+π.2^(−1) ln( (1+(1−x^2 )^(1/2) ((1−(1−x^2 )^(1/2) )^(−1) +β β=f(1)=∫_0 ^(π=) ln(1+cosθ)dθ=−πln(2) so ∫_0 ^π (1+xcosθ)dθ = πlnx +π2^(−1) ln((1+(1−x^2 )^(1/2^ ) )((1−(1−x^2 )^(1/2) )^(−1) −πln(2).


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 25821 by abdo imad last updated on 15/Dec/17

$a n s w e r t o q 25796 f_{} i n d t h e v a l u e o f f (x) = \int_{0^{}}^{π_{}} l n (1 + x c o s θ) d θ w i t h 0 < x < 1 \partial f / \partial x = \int_{0}^{π} c o s θ {(1 + x c o s θ)}^{- 1} d θ = π x^{- 1} - x^{- 1} \int_{0}^{π} {(1 + x c o s θ)}^{- 1} d θ a n d b y t h e c h a n g e e n t t a n θ = u t h e n t h e c h a n g e m e n t u = {((1 + x) {(1 + x)}^{- 1^{}})}^{} . t . w e f i n d \int_{0}^{θ} {(1 + x c o s θ)}^{- 1} d θ = π {(1 - x^{2_{}})}^{- 1 / 2}$ $s o \partial f / \partial x = π x^{- 1} - π x^{- 1} (1 - x^{- 1 / 2}) s o f (x) = π l n (x) - π \int^{x} t^{- 1} {(1 - t^{2})}^{- 1 / 2} d t$ $b u t \int^{x} t^{- 1} (1 - t^{2}) d t = - 1 . 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2} . {(1 - {(1 - x^{2})}^{1 / 2})}^{- 1})$ $a n d f (x) = π l n (x) + π . 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2} ({(1 - {(1 - x^{2})}^{1 / 2})}^{- 1} + β$ $β = f (1) = \int_{0}^{π =} l n (1 + c o s θ) d θ = - π l n (2) s o$ $\int_{0}^{π} (1 + x c o s θ) d θ = π l n x + π 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2^{}}) ({(1 - {(1 - x^{2})}^{1 / 2})}^{- 1} - π l n (2) .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum