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




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
$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\left({x}^{\mathrm{2}} +\mathrm{cos}\:{x}\right)\:\mathrm{log}\:\left(\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}\right){dx}\:=\:\mathrm{0} \\ $$
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
$${let}\:{put}\:\:{f}\left({x}\right)=\:\left({x}^{\mathrm{2}} +{cosx}\right){ln}\left(\frac{\mathrm{2}+{x}}{\left.\mathrm{2}−{x}\right)}\right)\:{we}\:{have} \\ $$$${f}\left(−{x}\right)\:=\left({x}^{\mathrm{2}} +{cosx}\right){ln}\left(\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}\right)\:=−\left({x}^{\mathrm{2}} +{cosx}\right){ln}\left(\frac{\mathrm{2}+{x}}{\left.\mathrm{2}−{x}\right)}\right) \\ $$$$=−{f}\left({x}\right)\:\:\:{f}\:{is}\:{impar}\Rightarrow\:\int_{−\mathrm{1}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\mathrm{0} \\ $$
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 .
$${I}=\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{cos}\:{x}\right)\mathrm{log}\:\left(\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}\right){dx} \\ $$$$\:\:=\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{cos}\:{x}\right)\mathrm{log}\:\left(\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}\right){dx} \\ $$$${using}\:\int_{{a}} ^{\:\:{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{\:\:{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$$\mathrm{2}{I}=\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{cos}\:{x}\right)\left[\mathrm{log}\:\left(\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}\right)+\mathrm{log}\:\left(\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}\right)\right]{dx} \\ $$$$\mathrm{2}{I}=\mathrm{0}\:\:\Rightarrow\:\:{I}=\mathrm{0}\:. \\ $$

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