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1-1-x-2-cos-x-log-2-x-2-x-dx-0-

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Question Number 27281 by kemhoney78@gmail.com last updated on 04/Jan/18
∫_(−1) ^1 (x^2 +cos x) log (((2+x)/(2−x)))dx = 0
∫1−1(x2+cosx)log(2+x2−x)dx=0
Commented by abdo imad last updated on 04/Jan/18
let put f(x)= (x^2 +cosx)ln(((2+x)/(2−x)))) we have f(−x) =(x^2 +cosx)ln(((2−x)/(2+x))) =−(x^2 +cosx)ln(((2+x)/(2−x)))) =−f(x) f is impar⇒ ∫_(−1) ^1 f(x)dx=0
letputf(x)=(x2+cosx)ln(2+x2−x))wehavef(−x)=(x2+cosx)ln(2−x2+x)=−(x2+cosx)ln(2+x2−x))=−f(x)fisimpar⇒∫−11f(x)dx=0
Answered by ajfour last updated on 04/Jan/18
I=∫_(−1) ^( 1) (x^2 +cos x)log (((2+x)/(2−x)))dx =∫_(−1) ^( 1) (x^2 +cos x)log (((2−x)/(2+x)))dx using ∫_a ^( b) f(x)dx=∫_a ^( b) f(a+b−x)dx 2I=∫_(−1) ^( 1) (x^2 +cos x)[log (((2+x)/(2−x)))+log (((2−x)/(2+x)))]dx 2I=0 ⇒ I=0 .
I=∫−11(x2+cosx)log(2+x2−x)dx=∫−11(x2+cosx)log(2−x2+x)dxusing∫abf(x)dx=∫abf(a+b−x)dx2I=∫−11(x2+cosx)[log(2+x2−x)+log(2−x2+x)]dx2I=0⇒I=0.

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