### Instruction: Find the area bounded by the graph of $y = \arccos(\sin x)$ and the $x$-axis on the interval $\frac{\pi}{2} \le x \le \frac{5 \pi}{2}.$.### Response: I need to find the area between a curve and the $x$-axis, so I will use the definite integral formula.
The area is equal to the integral from $\frac{\pi}{2}$ to $\frac{5 \pi}{2}$ of the function $\arccos(\sin x)$ with respect to $x$.
To evaluate this integral, I will use a substitution method.
I notice that the function $\arccos(\sin x)$ has a periodic pattern that repeats every $\pi$ units along the $x$-axis.
So I will let $u = x - \frac{\pi}{2}$ and $du = dx$.
This means that when $x = \frac{\pi}{2}$, $u = 0$ and when $x = \frac{5 \pi}{2}$, $u = 2 \pi$.
So the integral becomes the integral from $0$ to $2 \pi$ of $\arccos(\sin (u + \frac{\pi}{2}))$ with respect to $u$.
Now I can use the trigonometric identity $\sin (u + \frac{\pi}{2}) = \cos u$ to simplify the integrand.
The integral becomes the integral from $0$ to $2 \pi$ of $\arccos(\cos u)$ with respect to $u$.
But $\arccos(\cos u)$ is just $u$ for any $u$ between $0$ and $\pi$.
So the integral becomes the integral from $0$ to $\pi$ of $u$ with respect to $u$ plus the integral from $\pi$ to $2 \pi$ of $u$ with respect to $u$.
These are both simple integrals that can be evaluated by finding the antiderivative of $u$ and plugging in the limits of integration.
The antiderivative of $u$ is $\frac{u^2}{2} + C$.
So the first integral is $\frac{\pi^2}{2} - 0 = \frac{\pi^2}{2}$.
And the second integral is $\frac{(2 \pi)^2}{2} - \frac{\pi^2}{2} = \frac{3 \pi^2}{2}$.
So the area is the sum of these two integrals, which is $\frac{\pi^2}{2} + \frac{3 \pi^2}{2} = 2 \pi^2$.