Integrating an even function on a symmetric interval that equals to zero? So I find this integral very interesting since the function $f(x)=\frac{x^2-a^2}{\sqrt{x^2+a^2}}$ is an even function you expect the result to now be non-zero when you integrated it over a symmetric interval right?
$$\int_{-1}^{1}\frac{x^2-a^2}{\sqrt{x^2+a^2}}dx=0$$
But for this particular integral that is not the case,
First, rewrite the expression to integration by parts form that is set $u=x^2-a^2$ and $dv=\frac{1}{\sqrt{x^2+a^2}}dx$
From which $du=2xdx$ and $v=\tanh^{-1}\frac{x}{\sqrt{x^2+a^2}}$ rearranging the whole thing we get $$-2\int_{-1}^{1}\tanh^{-1}(\frac{x}{\sqrt{x^2+a^2}})xdx$$ the left-hand side vanished since integrating from -1 to 1 its equal to zero
Now since $-2\tanh^{-1}(\frac{x}{\sqrt{x^2+a^2}})=\ln(\frac{x}{\sqrt{x^2+a^2}}+1)-\ln(1-\frac{x}{\sqrt{x^2+a^2}})$
Which equals $$\ln\frac{\sqrt{a^2+x}-x}{\sqrt{a^2+x}+x}$$
Then, $$\int_{-1}^{1}\ln\frac{(\sqrt{a^2+x}-x)}{(\sqrt{a^2+x}+x)}xdx$$ which is simply if $s=x$ then $ds=dx$ which gives $$\int_{-1}^{1}\ln\frac{(\sqrt{a^2+s}-s)}{(\sqrt{a^2+s}+s)}ds$$ from this you can see easily that this equals to $\ln{(a^2)}$-$\ln{(a^2)}=0$
Now I’m wondering if there are many more functions like this?
 A: I didn't really go through your calculations (I suspect it only works for specific choices of $a$, or possibly not at all; double check all your calculations), but it is definitely possible for an even function to have zero integral over a symmetric interval. Consider a function $\phi:[0,\infty)\to\Bbb{R}$ (assumed RIemann-integrable over every compact interval... or for simplicity just say it is continuous), and suppose that for some $b>0$, we have $\int_0^b\phi(x)\,dx=0$.
Now, we define the function $f:\Bbb{R}\to\Bbb{R}$ as $f(x)=\phi(|x|)$. Then, $f$ is an even function, and
\begin{align}
\int_{-b}^bf(x)\,dx&=2\int_0^bf(x)\,dx\tag{$f$ is even}\\
&=2\int_0^b\phi(|x|)\,dx\\
&=2\int_0^b\phi(x)\,dx\\
&=0.
\end{align}
However, if you're dealing with positive even functions, then no, the integral is always going to be positive.
As a concrete example, take $f=0$, which obviously has zero integral. Another example is to take $f(x)=\cos x$, an even function which satisfies $\int_0^{\pi}\cos x\,dx=0$ (positive and negative parts up to $\pi/2$ cancel out), so $\int_{-\pi}^{\pi}\cos x\,dx=0$, and so on.
