Regarding whether two functions are even or odd I am solving the integral 
$$\int_\frac{-π}{2}^\frac{π}{2} x \cos x+ \tan x^5 \,dx$$
In the solution, both $x \cos x$ and $\tan x^5$ are said to be odd functions.
Are these functions even or odd?
I noticed:
$$\left(\frac{-π}{2} \right)\cos \left(\frac{-π}{2} \right)=-0$$  and $$\left(\tan \left(\frac{-π}{2} \right) \right)^5 =-∞$$
If these are odd functions then $0=-0$ and $∞=-∞$, the latter of which is making me have doubts.
 A: Note that odd/even-ness of a function must apply for all $x$ in the domain where the function is defined. Specific values don't necessarily mean anything in the grander context - to show a function is odd or even, it must be shown for arbitrary $x$.

Let $f(x) = x \cdot \cos(x)$. Consider $f(-x)$. Recalling cosine is an even function,
$$\begin{align}
f(-x) &=-x \cdot \cos(-x) \\
&= -x \cdot \cos(x) \\
&= -f(x) \\
\end{align}$$
Thus, $f$ is odd, as $f(-x) = -f(x)$.

Let $g(x) = \tan^5 (x)$. Recall tangent is odd, and thus
$$\begin{align}
g(-x) &= \tan^5 (-x)\\
&= (-1)^5 \tan^5 (x) \\
&= - \tan^5 (x) \\
&= - g(x)
\end{align}$$
(If instead you meant $\tan(x^5)$, the argument is pretty similar.)
Thus, $g$ is also odd.

It might also be worth noting:
You cannot say a function is "equal to" infinity. If $f(x) = 1/x$, $f(0)$ is not equal to $\infty$. In some contexts, it approaches infinity (as the right-hand limit as $x -> 0$), but $1/0 \neq \infty$.
This might be important to note when discussing even/odd-ness of such functions - the premise is that the function is defined in the first place. $\tan(x)$ is not defined for $x = k\pi$ for integers $k$. Not equal to infinity, or to negative infinity as might also be the case from a different viewpoint - just simply undefined.
A: To check, we have $$f(x)= x\cos(x)= x\cos(-x)= -(-x\cos(-x))= -f(-x)$$ So $f(x)=x\cos(x)$ is an odd function. Next, $$g(x)=(\tan(x))^{5}= \bigg(\frac{\sin(x)}{\cos(x)}\bigg)^{5} = \bigg(\frac{-\sin(-x)}{\cos(-x)}\bigg)^{5}=(-1)^{5}\bigg(\frac{sin(-x)}{\cos(-x)}\bigg)^{5}= -\bigg(\frac{sin(-x)}{\cos(-x)}\bigg)^{5}= -g(-x).$$ So we have the both are odd functions.
