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You need to remember how to exploit the symmetry of a function when evaluating the definite integral of this function.
- if f(x) is odd, then `int_-a^a f(x) dx = 0`
Since the problem claims that the definite integral of symmetric function is cancelling, then f(x) must be an odd function.
You need to remember the behaviour of an odd function:
`f(-x) = -f(x)`
You need to plug -x in the equation of the given function such that:
`f(-x) = ((-x)^2*sin(-2x))/((-x)^2+1)`
`f(-x) = (x^2*(-sin 2x))/(x^2+1) =gt f(-x) =- (x^2*(sin 2x))/(x^2+1) =gt f(-x) = -f(x)`
This proves that the function `f(x) = (x^2*(sin 2x))/(x^2+1)` is odd, hence `int_-a^a f(x) dx = 0` .
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