Integration Problem Proof  ($\sin x$) Problem:  Integration of $\displaystyle\int_{-1}^1 {\sin x\over 1+x^2} \; dx = 0 $
(according to WolframAlpha Definite Integral Calculator)
But I don't understand how. I tried to prove using integration by parts.
Here's the work:
$$
\int_{-1}^1 {\sin x\over 1+x^2} \; dx = \int_{-1}^1 {\sin x}{1\over 1+x^2} \; dx
$$
Let $u = \sin x,\quad du = \cos x\; dx\;$ and $v = \tan^{-1}x,\quad dv = {1\over 1+x^2}dx\;$.
So
$$
\int_{-1}^1 u dv = \left[uv\right]_{-1}^1 - \int_{-1}^1 v du =\left[ \sin x (\tan^{-1}x)\right]_{-1}^1 - \int_{-1}^1 \tan^{-1}x \cos x\; dx.
$$
Next let $u = \tan^{-1}x, du = {1\over 1+x^2}$ and $dv = \cos x, v = \sin x$...
I stopped here, because I feel like I'm going in a circle with this problem. What direction would I take to solve this because I don't know whether integration by parts is the way to go? Should I use trig substitution?
Thanks.

You don’t have to do any actual integration. Let $$f(x)=\frac{\sin x}{1+x^2}\;$$ then $$f(-x)=\frac{\sin(-x)}{1+(-x)^2}=\frac{-\sin x}{1+x^2}=-f(x)\;,$$ so $f(x)$ is an odd function. The signed area between $x=-1$ and $x=0$ is therefore just the negative of the signed area from $x=0$ to $x=1$, and the whole thing cancels out.
In more detail, let $$A=\int_0^1 f(x) dx=\int_0^1\frac{\sin x}{1+x^2} dx\;,$$ and let $$B=\int_{-1}^0 f(x) dx=\int_{-1}^0\frac{\sin x}{1+x^2} dx\;.\tag{1}$$ Now substitute $u=-x$ in $(1)$: $f(u)=f(-x)=-f(x)$, $du=-dx=(-1)dx$ so $dx=-du$, and $u$ runs from $1$ to $0$, so
$$B=\int_1^0 -f(x)(-1)dx=\int_1^0f(x)dx=-\int_0^1f(x)dx=-A\;.$$
Thus, $$\int_{-1}^1 f(x)dx=A+B=A-A=0\;.$$
Note that the specific function $f$ didn’t matter: we used only the fact that $f$ is an odd function.
