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?

The book is indeed incorrect. 
$$arccos(\frac{1}{3}) \neq \frac{\pi}{3}$$
Note if it was the case that $arccos(\frac{1}{3}) = \frac{\pi}{3}$ as stated in the solutions, 
then the integral would total 0. 
