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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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