Yes, your result is correct. For $x\in[-1,1]$,
$$\arccos(x)=\frac{\pi}{2}-\arcsin(x).$$
Hence
$$\int_0^{\pi/2}\arccos(\sin(x))dx=
\int_0^{\pi/2}\left(\frac{\pi}{2}-x\right)dx=\int_0^{\pi/2}tdt=\left[\frac{t^2}{2}\right]_0^{\pi/2}=\frac{\pi^2}{8}.$$
P.S. WA gives the correct result. Moreover $t\to \arccos(t)$ is positive in $[-1,1)$ so the given integral has to be POSITIVE!
The indefinite integral $\displaystyle \int f(x) \, d x$ represents the collection of all functions whose derivatives with respect to $x$ equals $f(x)$---they're antiderivatives.
This is in contrast with the definite integral $\displaystyle \int_a^b f(x) \, d x$ which represents the area under the curve $y = f(x)$ from $x = a$ to $x = b$.
These objects are related by the Fundamental theorem of calculus which says (in one of its many forms) that the area in the second expression can be computed by evaluating any one of the antiderivatives at the endpoints and subtracting.
Note that it does not say that using antiderivatives is the only way to compute the area under the curve, it's just a very convenient way in many situations. There are many situations in which it is impossible to write down a nice expression for the indefinite integral (e.g. in terms of elementary functions) but nevertheless possible to compute the area just fine.
One way of doing this is to exploit the geometry of the shape whose area we are after.
A classic example (mentioned in comments) is the Gaussian integral
$$ \int_{-\infty}^\infty e^{-x^2} \, d x = \sqrt{\pi}. $$
There are also examples where it is perfectly possible to find a nice antiderivative, but it may be difficult, or at least more difficult than some other approach.
For instance,
$$ \int_{-1}^1 \sqrt{1 - x^2} \, d x = \frac{\pi}{2}. $$
We could compute this by finding an antiderivative (which is doable but tricky if we're not familiar with trigonometric substitutions), but it is (probably) easier to notice that the shape we are after is really a semicircle. (You may argue that this is circular, depending on how you derived the formula for the area of a circle in the first place.)
As to why some functions don't have elementary antiderivatives: this is a fairly deep question to do with an area of mathematics called differential algebra. A reasonably approachable exposition I enjoy is Brian Conrad's Impossibility theorems for elementary integration.
Finally there are powerful methods to approximate definite integrals---to estimate the area under the curve $y = f(x)$---without having a clue whatsoever how one might find an antiderivative. This is an entire area of mathematics in its own right called numerical integration. Depending on what one is doing, this may be all one needs.
Best Answer
The first one gives $\frac {8-(-1)}3=3$. So the answers match. The anti-derivatives are both correct.