Find the value of the constant $A$ given the particle has the wavefunction $\psi(x,t=0)=A\cos^{3}\left(\frac{\pi x}{2a}\right)$ Consider a particle in an infinite square well with $$V(x)
\begin{cases}
= 0  &   −a \lt x \lt a\\
\to  \infty & \text{otherwise}
\end{cases}$$ 
At $t = 0$, the particle has the wavefunction (defined over $−a \lt x \lt a$):
$$\psi(x,t=0)=A\cos^{3}\left(\frac{\pi x}{2a}\right)$$
Find the value of the constant $A$.

Using De Moivre's theorem I was able to show that $$\psi(x,t=0)=A\cos^{3}\left(\frac{\pi x}{2a}\right)=\frac{3A}{4}\cos\left(\frac{\pi x}{2a}\right)+\frac{A}{4}\cos\left(\frac{3\pi x}{2a}\right)\tag{1}$$
& using the fact that $$\psi=\sum_na_n\phi_n\tag{2}$$ 
Using $\mathrm{(1)}$ & $\mathrm{(2)}$ I find that
$$\psi(x, t=0)=\sum_na_n\phi_n=a_1\phi_1+a_3\phi_3$$
so $a_1=\dfrac{3A}{4}$, $\,\,\phi_1=\cos\left(\dfrac{\pi x}{2a}\right)$, $\,\,a_3=\dfrac{A}{4}$, $\,\,\phi_3=\cos\left(\dfrac{3\pi x}{2a}\right)$
Now using 
$$\sum_{n=-\infty}^{\infty}{\lvert a_n \rvert}^2=1$$
So $$\left(\frac{3A}{4}\right)^2+\left(\frac{A}{4}\right)^2=1$$
Which, on rearranging, gives $$A=\sqrt{\frac85}$$

The problem is that the correct answer is $$A=\sqrt{\frac{8}{5a}}$$
I am very curious about why that factor of $a$ is part of the normalization constant $\mathrm{A}$ since it is also half the width of the well and appears in the argument of the cosine eigenstate. How can this normalization constant have a property of the system?
I have applied what I thought was the correct logic. But it seems I am missing something. Does anyone have any idea how the author reached that answer?

EDIT:
I have been given two answers to this question, thank you to those that took the time to answer. 
There is still one thing I can't understand, and unfortuanately I will have to upload the full question and solution in order to get the point across. Below is a 2nd year undergraduate physics assessed problem:


Here is the professors solution to the first part of the question:


and here is the solution to the second part of the question:

Many thanks.
 A: You're working far too hard.  Just normalize your wave function:
$$\int\limits_{-a}^a \left| A \cos^3 \left( {\pi x \over 2 a} \right) \right|^2 dx = A^2 \int\limits_{-a}^a \cos^6 \left( {\pi x \over 2 a}\right) dx = 1$$
where the integral is computed symbolically instantly in Mathematica or can be found in a table or can be computed through the trigonometric substitution given by Omnomnomnom, below.
Regardless, this method gives $A = \sqrt{{8 \over 5 a}}$ very quickly indeed.
Incidentally, indeed your (incorrect) answer must depend upon $a$.  Qualitatively, if $a$ is large (the normalized wave function is spread out), then its amplitude $A$ must be small.  In fact, if you think about the physics, the the amplitude must go down as $1/\sqrt{a}$.
A: I would like to state that you have a mistake.  The (normalized) eigenstate function $\phi_n$ should look like
$$\phi_n(x)=\frac{1}{\sqrt{a}}\,\cos\left(\frac{n\pi x}{2a}\right)\text{ for }x\in[-a,+a]\,.$$
This can be easily checked since
$$\int_{-a}^{+a}\,\cos^2\left(\frac{n\pi x}{2a}\right)\,\text{d}x=a\,.$$
Therefore, from (1) and (2) in your question, you should get $$a_1=\frac{3A\sqrt{a}}{4}\text{ and }a_3=\frac{A\sqrt{a}}{4}\,.$$
