How to solve $\int^1_{-1} \frac{\sin(x)}{1+x^2}dx$?

Since, $\frac{\sin(-x)}{1+(-x)^2}=-\frac{\sin(x)}{1+x^2}$ the function is symmetric over the interval of $[-1,1]$. Therefore we can evaluate the integral as:

$$\int^1_{-1} \frac{\sin(x)}{1+x^2}\,dx=0$$

Here is a visual representation of what I mean:

Hint: Try $u = -x$. You may get something very similar but different. Actually, this substitution will give $I = -I$, if $I$ is the original integral.
You can solve that integral when the integration interval is given as $[-a, a]$.

HINT

Since $\sin(-x)=-\sin(x)$ you do not need any substitution at all. What do you know about an integral with symmetric boundaries of an odd function?

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For a given odd function $f(x)$, i.e. $f(-x)=-f(x)$, integrated over a symmetric interval $[-a,a]$, note that we get the following by enforcing the substitution $x\mapsto -x$

\begin{align*} \int_{-a}^af(x)\mathrm dx&=\int_a^{-a}f(-x)(-1)\mathrm dx\\ &=\int_{-a}^af(-x)\mathrm dx\\ &=-\int_{-a}^af(x)\mathrm dx\\ \therefore~2\int_{-a}^af(x)\mathrm dx&=0\\ \end{align*}

$$\therefore~\int_{-a}^af(x)\mathrm dx~=~0$$

Now consider the function $f(x)=\frac{\sin x}{1+x^2}$. Is this one odd; if so what is the integral over the interval $[-1,1]$?