# Why the definite integral of the function x^2*sin2x/(x^2+1) is 0? x=-3 to x=3

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We find the definite integral of the function x^2*sin2x/(x^2+1) is x=-3 to x=3.

Let f(x) = x^2*sin2x/(x^2+1.)

Then f(-x) = (-x)^2*sin(2*-x)/{(-x)^2+1)} = x^2 (-sin2x)/(x^2+1) = f(-x).

Therefore Integral Int {f(x) dx from x= 0 to x = a } = A.

Then Int f(x)dx from x = -a to x= 0 = {Int f(x)dx from x= 0 to x = -a } = -A.

Therefore Int f(x) dx = A+(-A) = 0.

The answer is very simple one. Because the given function is continuous and odd.

But let's see how it works. We know that if a function is continuous over a range [-a ; a], then the following identity is true:

Int f(x)dx(-a --> a) = Int [f(x) + f(-x)]dx ( 0--> a)

Furthermore, if the function is odd, then the above identity is changing in:

Int f(x)dx(-a --> a) = 0

A function is odd if and only if f(-x) = -f(x).

We'll verify if the given function is odd. We'll substitute x by -x:

f(-x) = (-x)^2*sin2(-x)/[(-x)^2+1]

f(-x) = x^2*sin (-2x)/(x^2 + 1)

Since sine function is odd, then sin (-2x) = - sin 2x.

f(-x)= -x^2*sin (2x)/(x^2 + 1)

**f(-x) = -f(x)**

**So, the given function is odd, then :**

**Int [x^2*sin2x/(x^2+1)]dx (-3 --> 3) = 0**