Integration substitution rule I am trying to evaluate
$$\int_{-\pi/4}^{\pi/4}(x^3+x^4\tan x)\,dx.$$
I tried to factor out an $x^3$ but that did nothing for me, and I also attempted to split it up into $x^3$ and $x^4 \tan x$ but that didn't help and I am not sure if it is legal. Nothing really seems to be quite right for this.
 A: Clearly the integral is zero being an integral of an odd function over a symmetric domain:
$$
  \int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan(x) \right) \mathrm{d} x \stackrel{\color\maroon{x \to -x}}{=}  -\int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan(x) \right) \mathrm{d} x
$$
A: $$\int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan x \right) dx  = \int_{-\pi/4}^{\pi/4} x^3 dx + \int_{-\pi/4}^{\pi/4}x^4 \tan x$$
If $f(x)=x^3$ then clearly $f(-x)=-f(x)$ (i.e. it is an odd function).  Thus, because the absolute value of the bounds are equal
$$\int_{-\pi/4}^{\pi/4} x^3 dx = 0$$
Letting $g(x)=x^4, h(x)=\tan x$, we have $g(-x)=g(x)$ and $h(-x)=-h(x)$.  Thus
$$g(-x)h(-x)=-g(x)h(x)$$
confirming that $x^4\tan x$ is odd as well, so 
$$\int_{-\pi/4}^{\pi/4}x^4 \tan x=0$$
Confirming that
$$\int_{-\pi/4}^{\pi/4} \left(x^3 + x^4 \tan x \right) dx =0$$
A: The function $f(x) = x^3 + x^4\tan(x)$ i odd. We check this by replacing the $x$ by a $-x$:
$$f(-x) = (-x)^3 + (-x)^4\tan(-x) = -x^3 + x^4\cdot(-\tan(x)) = -f(x).$$
We have used that $\tan$ is an odd function. 
Hence you have an integral over an interval symmetric around zero, and so the integral is zero.
