# 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.

• the fact that $\left(\tan \left(\frac{-π}{2} \right) \right)^5 =-∞$ (...suitably interpreted) is evidence that the function is indeed odd, since $\left(\tan \left(\frac{+π}{2} \right) \right)^5 =∞$ – Calvin Khor Jan 21 at 12:19

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.

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.