Evaluating the integral $\int_{-1}^{1} \frac{\sin{x}}{1+x^2}dx$ I was asked to evaluate the integral  
$$\int_{-1}^{1} \frac{\sin{x}}{1+x^2}dx$$
if it exists.
This is a problem from Calculus and the student has been taught how to use trigonometric substitution.  My intuition was to do trig sub with $$x=\tan{\theta}$$ and eliminating $$\frac{dx}{1+x^2}$$ but when I do that the numerator becomes $$\sin{(\tan{\theta})}$$ which I am not comfortable integrating.
Another queue was that the problem asks "if it exists" so I tried to see if there are any points within $(-1,1)$ that may cause any problem, but I don't think I see any asymptotes or undefined numbers so I'm not sure if I am dealing with an improper integral.
Can someone help me out on this? Thank you.
 A: The integral is
\begin{align}
I &= \int_{-1}^{1} \frac{\sin(x)}{1+x^{2}} dx \\
&= \int_{-1}^{0} \frac{\sin(x)}{1+x^{2}} dx + \int_{0}^{1} \frac{\sin(x)}{1+x^{2}} dx \\
&= - \int_{0}^{1} \frac{\sin(x)}{1+x^{2}} dx + \int_{0}^{1} \frac{\sin(x)}{1+x^{2}} dx \\
&= 0 
\end{align}
where the change of variable $x \rightarrow -x$ was made. 
A: We will begin by evaluating $$I(x) = \int\!\dfrac{\sin(x)}{1+x^2}\mathrm{d}x$$
$$I  = \dfrac{1}{2i} \int\!\dfrac{\sin(x)}{x-i}\mathrm{d}x -  \dfrac{1}{2i}\int\!\dfrac{\sin(x)}{x+i}\mathrm{d}x$$
$$I  = \dfrac{1}{2i} \int\!\dfrac{\sin(x)}{x-i}\mathrm{d}x -  \dfrac{1}{2i}\int\!\dfrac{\sin(x)}{x+i}\mathrm{d}x$$
We will now consider $$J_\pm = \int\!\dfrac{\sin(x)}{x\pm i}\mathrm{d}x$$
Substitute $u = x \pm i$.
$$J_\pm = \int\!\dfrac{\sin(u \mp i)}{u}\mathrm{d}x = \int\!\dfrac{\sin(u)\cos(i)\mp\sin(i)\cos(u)}{u}\mathrm{d}x$$
Splitting the integral and using hyperbolic functions
$$J_\pm = \int\!\dfrac{\sin(u)\cosh(1)}{u}\mathrm{d}x \mp\int\!i\dfrac{\sinh(1)\cos(u)}{u}\mathrm{d}x$$
Writing in terms of $\operatorname{Si}(x)$ and $\operatorname{Ci}(x)$
$$J_\pm = \cosh(1)\operatorname{Si}(u) \mp i\sinh(1)\operatorname{Ci}(u)$$
$$J_\pm = \cosh(1)\operatorname{Si}(x\pm i) \mp i\sinh(1)\operatorname{Ci}(x\pm i)$$
$$I = \dfrac{1}{2i}\left(J_--J_+\right)$$
$$I = \dfrac{\left(\cosh(1)\operatorname{Si}(x- i) + i\sinh(1)\operatorname{Ci}(x- i)\right)-\left(\cosh(1)\operatorname{Si}(x+ i) - i\sinh(1)\operatorname{Ci}(x+ i)\right)}{2i}$$
Now evaluate $I$ at $1$ and $-1$. 
$$I(1) = I(-1) = -0.324967578038053553$$
$$I(1) - I(-1) = \boxed{0}$$
If you are sneaky you can use symmetry to show that $I(1) = I(-1)$ without actually calculating the value.
A: Zero by symmetry.  
It is an odd integrand integrated over a symmetric bound
or more properly "integrated over an interval that is symmetric about the origin".
