Evaluate $\int_{-1}^{1} \frac{x^3\,e^{\sin \left( \cos \left( x \right) \right)}+1}{x^2+1}\mathrm{d}x$ 
$$\int_{-1}^{1} \frac{x^3\,e^{\sin \left( \cos \left( x \right) \right)}+1}{x^2+1}\mathrm{d}x$$

$\cos x$ is an even function, $\sin(even)$ is a compostion of of an odd function and an even function which is an even function.
$e^{x}$ is neither even nor odd, so the function $e^{\sin( \cos ( x ))}$ is even, now $x^3\,e^{\sin \left( \cos \left( x \right) \right)}$ is a product of an even and odd function so it is odd.
Overall we got $\int_{-1}^{1} \frac{odd+1}{even+1}$ 
What Can we say about $\int_{-1}^{1} \frac{odd+1}{even+1}=\int_{-1}^{1} \frac{odd}{even}$?

$I=\displaystyle\int_{-1}^{1} \frac{x^3\,e^{\sin \left( \cos \left( x \right) \right)}+1}{x^2+1}\,dx$
$=\displaystyle\int_0^1\left (\frac{x^3\,e^{\sin \left( \cos \left( x \right)) \right)}+1}{x^2+1}+\frac{-x^3\,e^{\sin \left( \cos \left( x \right) \right)}+1}{x^2+1}\right )\,dx$  $\quad (*)$
$=\displaystyle\int_0^1 \frac{2}{x^2+1}\,dx=\frac{\pi}{2}$

$(*)\to \quad  \int_{-a}^{a}f(x)\,dx=\int_0^a (f(x)+f(-x))\,dx$ 
(Simply write $\int_{-a}^{a}f=\int_{-a}^0f+\int_0^af$ and then substitute $x=-t$ in the 1st integral.)
