How can I calculate $\int_{-5}^5 \dfrac{x^3 \sin^2x}{x^4+2x^2+1}~dx$?

Question

I was wondering what kind of approach could be used for this integration: $$\int_{-5}^5 \dfrac{x^3 \sin^2x}{x^4+2x^2+1}~dx$$

Answer 1

HINT:

Observe that the function is odd and the interval $[-5,5]$ is symmetric about $0$.

See more in this question:

Definite integral of an odd function is 0 (symmetric interval)

Answer 2

$$ x^4 + 2x^2 + 1 = \left(x^2+1\right)^2 $$ so we have an integral of the form $$ \int \frac{x^3}{\left(x^2+1\right)^2}\sin^2 x dx $$ we can clearly see that $$ f(x) = \frac{x^3}{\left(x^2+1\right)^2}\sin^2 x \implies f(-x) = -f(x) $$ i.e. an Odd function $$ \int_{-5}^5f(x)dx = \int_{-5}^0f(x)dx + \int_{0}^5 f(x)dx = \int_{5}^0 f(-x)(-dx) + \int_{0}^5 f(x)dx = \int_0^5f(-x)dx + \int_{0}^5 f(x)dx $$ now the final bit is to use $f(-x) = -f(x)$ we obtain $$ \int_{-5}^5f(x)dx = -\int_0^5f(x)dx + \int_{0}^5 f(x)dx = 0 $$

Answer 3

Once you notice it's an odd function,the answer will follow.

definite integral of odd function