Integration changing limits - does the question have an error? I am asking about changing the limits of integration. 
I have the following integral to evaluate - 
$$\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}dx$$ using the substitution $x = sec \theta$. 
The problem states
Use the substitution to change the limits into the form $\int_a^b$ where $a$ and $b$ are multiples of $\pi$.
Now, this is what I did. 
$$ x= \sec \theta$$
$$\frac{dx}{d\theta} = \sec\theta \tan\theta$$
$$dx = sec\theta tan\theta \ d\theta$$
$$\begin{align}\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}\,dx \\
&= \int\frac{1}{(\sec^2\theta-1)^{\frac{3}{2}}}\sec\theta \tan\theta \,d\theta \\
&= \int\frac{\sec\theta \tan\theta}{(\tan^2\theta)^{\frac{3}{2}}} \,d\theta \\
&= \int\frac{\sec\theta \tan\theta}{\tan^3\theta} \, d\theta \\
&= \int\frac{\sec\theta}{\tan^2\theta} \, d\theta \\
&= \int\frac{\cos\theta}{\sin^2\theta} \, d\theta \\
&= \int \csc\theta \cot\theta \, d\theta
\end{align}$$
But here is my problem. 
I know that when $x = 2$, 
$$2 = \sec \theta$$
$$\frac{1}{2} = \cos \theta$$
$$\frac{\pi}{3} = \theta$$
but when $x = 3$
$$3 = \sec \theta$$
$$\frac{1}{3} = \cos \theta$$
$$\arccos\left(\frac{1}{3}\right) = \theta = $$
but this does not give me a definite result in $\pi$. 
The book says the following - 

where $\arccos\left(\frac{1}{3}\right) = \frac{\pi}{3}$
Am I the only one or is the book wrong in this instance?

For simplicity, write $a=\arccos\frac{1}{2}$ and $b=\arccos\frac{1}{3}$. One can take care of them at the end.
It is true that $a=\pi/3$, but it's definitely wrong that $b=\pi/3$. Actually, $b\approx1.230959$, but you won't need that.
The integral becomes
$$
\int_a^b-\frac{1}{(\sec^2\theta-1)^{3/2}}\sec^2\theta\sin\theta\,d\theta
$$
Now it's better to write the integrand in terms of sine and cosine, recalling that the interval of integration is contained in $(0,\pi/2)$. Thus
$$
\sec^2\theta-1=
\frac{1-\cos^2\theta}{\cos^2\theta}=
\frac{\sin^2\theta}{\cos^2\theta}
$$
so the integrand is
$$
-\frac{\cos^3\theta}{\sin^3\theta}\frac{1}{\cos^2\theta}\sin\theta=
-\frac{\cos\theta}{\sin^2\theta}
$$
Thus we get
$$
\int_2^{3}\frac{1}{(x^2-1)^{3/2}}\,dx=
\int_a^b-\frac{\cos\theta}{\sin^2\theta}\,d\theta=
\int_a^b-\frac{1}{\sin^2\theta}\,d(\sin\theta)=
\left[\frac{1}{\sin\theta}\right]_a^b
$$
Since $a=\arccos(1/2)$ and $b=\arccos(1/3)$, we get
$$
\sin a=\sqrt{1-\cos^2a}=\frac{\sqrt{3}}{2}
\qquad
\sin b=\sqrt{1-\cos^2b}=\frac{2\sqrt{2}}{3}
$$
with no “sign uncertainty”, because $a$ and $b$ lie in $(0,\pi/2)$.
Thus the integral is
$$
\frac{2\sqrt{2}}{3}-\frac{\sqrt{3}}{2}
$$
