# How do I solve $\int_{-5}^{5} \frac{x^3 \sin^2x}{x^4 +2x^2+1}\,dx$?

$$\int_{-5}^{5} \frac{x^3 \sin^2x}{x^4 +2x^2+1}\,dx$$

I only have basic calculus but this is what I tried.

Firstly this can't be integrated directly.

I tried to make a substitution, letting $u$ equal to varies parts of this expression but to no avail.

I tried integration by parts but it just got too messy.

I did notice that $x^4+2x^2+1 = (x^2+1)^2$ but that didn't get me anywhere either.

I did plot this in Geogebra and noticed that this function is origin-symmetric so $2\int_{0}^{5}f(x)\,dx$ could be used to simplify things after the integration but it doesn't help to do the actual integration.

How does one go about tackling this particular integral?

• Hint: "it's a nice round number". – Jack D'Aurizio Jun 27 '17 at 20:48
• $\color{#f00}{\Huge 0}$. – Felix Marin Jun 28 '17 at 4:54

Hint: Let the integrand be $f(x)$.

We have that $f(-x)=-f(x)$, seen by the term $x^3$.

The integral of an odd function from $-a$ to $a$ is $0$.

• Probably you mean $f(x)=-f(\color{red}-x)$. – Jaideep Khare Jun 27 '17 at 20:20
• @JaideepKhare Multiply both sides of your equation by $-1$. – projectilemotion Jun 27 '17 at 20:20
• @projectilemotion Ohh! I am so dumb. My brain was on vacation writing that. – Jaideep Khare Jun 27 '17 at 20:22
• @JaideepKhare I can relate lol – mrnovice Jun 27 '17 at 20:22
• my brain's still on vacation @JaideepKhare – Saketh Malyala Jun 27 '17 at 20:23

HINT

It is an odd function, so the integration should be 0

The function under integral is odd so

$$\\ \int _{ -5 }^{ 5 } \frac { { x }^{ 3 }\sin ^{ 2 }{ x } }{ x^{ 4 }+2x^{ 2 }+1 } dx=\int _{ -5 }^{ 5 } \frac { { x }^{ 3 }\sin ^{ 2 }{ x } }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 } } dx=0$$

Let $f(x)=\frac{x^3\sin^2x}{(x^2+1)^2}$.

Since $f(-x)=-f(x)$, we obtain: $$\int_{-5}^5f(x)dx=\int_{-5}^0f(x)dx+\int_{0}^5f(x)dx=$$ $$=\int_{-5}^0\left(-f(-x)\right)dx+\int_{0}^5f(x)dx=$$ $$=\int_{0}^5f(-x)dx+\int_{0}^5f(x)dx=$$ $$=\int_{0}^5\left(f(-x)+f(x)\right)dx=0.$$