int_{ - 3}^3 {frac{{{x^2}sin 2x}}{{{x^2} + 1} | vedclass

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(A) Let $f(x) = \frac{{{x^2}\sin 2x}}{{{x^2} + 1}}$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \frac{{{{( - x)}^2}\sin (2( -

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(A) Let $f(x) = \frac{{{x^2}\sin 2x}}{{{x^2} + 1}}$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \frac{{{{( - x)}^2}\sin (2( - x))}}{{{{( - x)}^2} + 1}} = \frac{{{x^2}\sin ( - 2x)}}{{{x^2} + 1}} = \frac{{ - {x^2}\sin 2x}}{{{x^2} + 1}} = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{ - a}^a {f(x)\,dx = 0}$.Therefore,$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = 0}$.

$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

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