TITLE: Finding the integral of a trig function using a matrix
QUESTION [0 upvotes]: It can be shown that Ɓ = {1, $\cos(t)$,…$\cos(6t)$ and Ƈ = (1,$\cos(t)$,…$\cos^6$(t)} span the same subspace of Ƈ(ℝ)
a. Use an appropriate change of coordinate matrix to find $cos^6$(t) in terms of the basis Ɓ. Give the matrix  you used, and show the matrix multiplication you used to find the result.
b. Use your work in part (a) to help find ∫ $\cos^6$(t) dt. Give your answer as a linear combination of the elements of Ɓ.
I have done part a. but need help with part b. Here is the work I have so far:
First I construct a change of coordinate matrix using the identities I found from Maple:
P$_{B←C}$ =
\begin{bmatrix}
1 & 0 & 1/2 & 0 & 3/8 & 0 & 5/16 \\
0 & 1 & 0 & 3/4 & 0 & 5/8 & 0 \\
0 & 0 & 1/2 & 0 & 1/2 & 0 & 15/32 \\
0 & 0 & 0 & 1/4 & 0 & 5/16 & 0 \\
0 & 0 & 0 & 0 & 1/8 & 0 & 3/16 \\
0 & 0 & 0 & 0 & 0 & 1/16 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1/32 \\
\end{bmatrix}
To find ∫ $cos^6$(t) dt in terms of Ɓ:
[x]$_Ɓ$ =  P$_{B←C}$ [x]$_Ƈ$ where [x]$_Ƈ$ = 
\begin{bmatrix}
0 \\
0 \\
0 \\
0 \\
0 \\
0 \\
1 \\
\end{bmatrix}
When you multiply the two matrices you get [x]$_Ɓ$:
\begin{bmatrix}
5/16 \\
0 \\
15/32 \\
0 \\
3/16 \\
0 \\
1/32 \\
\end{bmatrix}
This corresponds to: 5/16 + 15/32 cos(2t) + 3/16 cos(4t) + 1/32 cos(6t).
Now I am not too sure how to go about part b.

REPLY [0 votes]: You just showed that $\cos^6 t = \dfrac{5}{16} + \dfrac{15}{32}\cos(2t) + \dfrac{3}{16}\cos(4t) + \dfrac{1}{32}\cos(6t)$. 
Using a basic $u$-substitution, we have $\int \cos(at)\,dt = \dfrac{1}{a}\sin(at)+C$. Hence: 
$\displaystyle \int \cos^6 t \,dt = \int \left(\dfrac{5}{16} + \dfrac{15}{32}\cos(2t) + \dfrac{3}{16}\cos(4t) + \dfrac{1}{32}\cos(6t)\right)\,dt$ $\dfrac{5}{16}t + \dfrac{15}{64}\sin(2t) + \dfrac{3}{64}\sin(4t) + \dfrac{1}{192}\sin(6t) + C$. 
The problem here is that every function in $B$ is bounded, and thus, any linear combination of elements in $B$ will also be bounded. However, the antiderivative of $\cos^6 t$ that we found above is not bounded as $t \to \infty$. Therefore, we cannot express it as a linear combination of elements in $B$.
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