Mathematics

$$\displaystyle\int _{ -5 }^{ 5 }{ \dfrac { { x }^{ 5 } { sin }^{ 2 } x }{ 1 +{ x }^{ 2 } + x^{ 4 } }  } dx$$ is equal to


SOLUTION
$$I =\displaystyle\int_{-5}^{5} \dfrac{x^{5}sin^{2}x}{1+x^{2}+x^{4}}dx $$

As we know that $$\displaystyle\int_{-a}^{a} f(x) = 0 $$ if $$ f(1-x) = -f(x) $$

So,
$$ f(x) = \dfrac{x^{5}sin^{2}x}{1+x^{2}+x^{4}} $$

$$ f(-x) = \dfrac{(-x)^{5}sin^{2}(-x)}{1+(-x)^{2}(-x)^{4}} = \dfrac{-x^{5}sin^{2}x}{1+x^{2}+x^{4}} $$

$$ = -f(x) $$

$$ \therefore I = 0 $$ 
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Subjective Medium Published on 17th 09, 2020
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