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Question Number 7932 by tawakalitu last updated on 24/Sep/16

Commented by sou1618 last updated on 25/Sep/16

set  f(x)=((sinx)/(1+x^2 +x^4 ))  f(−x)=((sin(−x))/(1+(−x)^2 +(−x)^4 ))=−f(x)  so  f(−x)+f(x)=0  ⇔∫_0 ^1 f(−x)+f(x)dx=0  ⇔∫_0 ^(−1) ((−sinx)/(1+x^2 +x^4 ))dx+∫_0 ^1 ((sinx)/(1+x^2 +x^4 ))dx=0  ⇔−∫_(−1) ^0 ((−sinx)/(1+x^2 +x^4 ))dx+∫_0 ^1 ((sinx)/(1+x^2 +x^4 ))dx=0  ⇔∫_(−1) ^1 ((sinx)/(1+x^2 +x^4 ))dx=0  ⇔∫_(−1) ^1 ((sinx)/(1+x^2 +x^4 ))dx=0

$${set}\:\:{f}\left({x}\right)=\frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} } \\ $$$${f}\left(−{x}\right)=\frac{{sin}\left(−{x}\right)}{\mathrm{1}+\left(−{x}\right)^{\mathrm{2}} +\left(−{x}\right)^{\mathrm{4}} }=−{f}\left({x}\right) \\ $$$${so} \\ $$$${f}\left(−{x}\right)+{f}\left({x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left(−{x}\right)+{f}\left({x}\right){dx}=\mathrm{0} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{−\mathrm{1}} \frac{−{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=\mathrm{0} \\ $$$$\Leftrightarrow−\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{−{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=\mathrm{0} \\ $$$$\Leftrightarrow\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=\mathrm{0} \\ $$$$\Leftrightarrow\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx}=\mathrm{0} \\ $$

Commented by tawakalitu last updated on 25/Sep/16

Thanks so much sir.

$${Thanks}\:{so}\:{much}\:{sir}. \\ $$

Answered by prakash jain last updated on 02/Oct/16

answer in comments

$$\mathrm{answer}\:\mathrm{in}\:\mathrm{comments} \\ $$

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