# Applying Parseval's relation to finding an integral

1. Jan 27, 2013

### SithsNGiggles

1. The problem statement, all variables and given/known data

Determine the result of the integral without using the Fundamental Theorem of Calculus:
$\displaystyle\int_{-\pi}^{\pi} \sin^2{t} \cos{2t} dt$,​
given the orthonormal basis $\left\{ \dfrac{1}{\sqrt{2}}, \cos{t}, \cos{2t} \right\}$.

2. Relevant equations

Parseval's relation, as per my lecture notes:
Let $V$ be an inner product space with an orthonormal basis $\{ x_1, x_2, \ldots, x_n \}$. Then $\forall u,v \in V$, with $u=\displaystyle \sum_{i=1}^n a_i x_i$ and $v=\displaystyle \sum_{i=1}^n b_i x_i$, we have
$\langle u,v \rangle = \displaystyle \sum_{i=1}^n a_i \overline{b_i}$, where
$a_i = \langle u, x_i\rangle$ and $b_i = \langle v,x_i\rangle$.

The inner product we were using in this particular example is defined by $\langle f,g\rangle := \dfrac{1}{\pi} \displaystyle \int_{-\pi}^{\pi} f(t)g(t) dt$.

3. The attempt at a solution

This was an example done during class, but it was done quickly just as class was ending.

I can see that $\left\langle \sin^2{t}, \cos{2t} \right\rangle = \dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} \sin^2{t} \cos{2t} dt$, so the integral itself will be $\pi \left\langle \sin^2{t}, \cos{2t} \right\rangle$.

So, I clearly have to determine the inner product $\left\langle \sin^2{t}, \cos{2t} \right\rangle$.

$\sin^2{t} \in \mbox{span}\left\{ \dfrac{1}{\sqrt{2}}, \cos{t}, \cos{2t} \right\}$, since $\sin^2{t}$ can be written as the linear combination $\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{2}} + \left(-\dfrac{1}{2}\right) \cos{2t}$.

I've tried recreating what the instructor presented:
$\left\langle \sin^2{t},\cos{2t} \right\rangle = \left\langle \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{2}} + \left(-\dfrac{1}{2}\right) \cos{2t}, \cos{2t} \right\rangle$

(Skipping some steps, and applying the linearity property of inner product)
$\left\langle \sin^2{t},\cos{2t} \right\rangle = \dfrac{1}{\sqrt{2}}\left\langle \dfrac{1}{\sqrt{2}}, \cos{2t} \right\rangle + \left(-\dfrac{1}{2}\right) \left\langle\cos{2t}, \cos{2t} \right\rangle$

$\left\langle \sin^2{t},\cos{2t} \right\rangle = 0 + \left(-\dfrac{1}{2}\right)$

From this, we get
$\displaystyle\int_{-\pi}^{\pi} \sin^2{t} \cos{2t} dt = -\dfrac{\pi}{2}$.

I've looked over this multiple times, but I can't seem to see where Parseval's relation comes into play. It could be that I've recreated the wrong thing and found some other way to find the answer. Any ideas where I might have gone wrong? Thanks