Definite integral (mixture of functions) I have problem with this integral and I generally don't know how to approach it:
$$\int_{-2}^2 (x^4+4x+\cos(x))\cdot \arctan\left(\frac{x}{2}\right)dx$$
I know that I probably have to make some substitution, but really don't know which.
Thanks for your help.

HINT:
For even function $f(x)$, $$\int_{-a}^af(x)dx=2\int_0^af(x)dx$$
For odd  $$\int_{-a}^af(x)dx=0$$
$$(-x)=x^4,\cos(-x)=\cos x,\arctan\left(-\dfrac x2\right)=-\arctan\dfrac x2$$
$\implies$ if $\displaystyle g(x)=\left(x^4+4\cos x\right)\arctan\frac x2,g(-x)=-g(x)$
If $\displaystyle h(x)=x\cdot\arctan\dfrac x2,h(-x)=h(x)$
So, the given integral reduces to $$2\int_0^24x\cdot\arctan\dfrac x2 dx$$
