When I use the calculator to get the result of this integral I get a decimal number. However, in my calculus book, as shown on the picture, the answer was represented using some trig symbols!
Is there is a way to show the same result, but not in decimal only? I want to use symbols like pi or sin to represent my answer. Just like my book. could it be done using the calculator?
Your calculator won't be able to do this in general. However, if you have an expression of the form $\tan^{-1}(\tan y)$, this can be simplified using facts about the tangent function.
First of all, let $t=\tan y$. There are infinitely many values of $x$ such that $\tan x = t$ (because the tangent function is periodic). Since the period of $\tan x$ is $\pi$, that means if you have one angle $x$ whose tangent is $t$, then $\tan(x+k\pi)=t$ as well; in fact, these are all such values.
Now, the arctangent (inverse tangent) function $\tan^{-1}(t)$ chooses a number $x$ such that $\tan x=t$ and $\displaystyle-{\pi\over2}\le x \le {\pi\over2}$; this value exists and is unique. So you need to find the value of $k$ that puts $x+k\pi$ into this interval.
For example, let $t=\tan9$. Then $9$ is one of the reals $x$ such that $\tan x=\tan 9$. The rest are $9+k\pi$. Now you need to choose $k$ so that $\displaystyle-{\pi\over2}\le 9+k\pi$ and $\displaystyle 9+k\pi \le {\pi\over2}$. The first inequality, when solved for $k$, yields $\displaystyle k \ge -{1\over2}\cdot{\frac {18+\pi }{\pi }}=-3.3647\ldots$. The second inequality, when solved for $k$, yields $\displaystyle k\le {1\over2}\cdot{\frac {\pi -18}{\pi }}=-2.3647\ldots$. Since $k$ must be an integer, $k=-3$, and so $\tan^{-1}(\tan 9)=9+(-3)\pi=9-3\pi$.