Integration (inner product)

• Dustinsfl

Dustinsfl

Inner product:

$\displaystyle <f,g>=\frac{1}{\pi}\int_{-\pi}^{\pi}fg \ dx=\begin{cases}0 & \ \text{if} \ f=g\\1 & \ \text{if} \ f\neq g\end{cases}$

Basis:
$\displaystyle\left\{\frac{1}{\sqrt{2}},\cos\theta, \sin\theta,\cdots\right\}$

I am trying to remember how to integrals of the form:

$\displaystyle \int_{-\pi}^{\pi}\sin^{a}\theta\cos^b (2\theta) \ d\theta$

However, I getting no where.

I left some guidance with these two integrals and I should be good to go then.

$\displaystyle\int_{-\pi}^{\pi}\sin^4\theta \ d\theta$

$$\Rightarrow\int_{-\pi}^{\pi}\left(\frac{1}{2}-\frac{\cos(2\theta)}{2}\right)^2 \ d\theta$$

$$\Rightarrow \int_{-\pi}^{\pi}\left(\left(\frac{1}{\sqrt{2}}\right)^2-\frac{\cos(2\theta)}{2}\right)^2 \ d\theta$$

Now, I am drawing a blank.

The other one I need guidance on is:

$\displaystyle\int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta$

$$\Rightarrow\int_{-\pi}^{\pi}\left(\left(\frac{1}{\sqrt{2}}\right)^2-\frac{\cos(2\theta)}{2}\right)^2\cos(2\theta) \ d\theta$$

$$\Rightarrow\int_{-\pi}^{\pi}\frac{\cos(4\theta)\cos(2\theta)}{4} \ d\theta$$

Now I am stuck again.

Last edited:
The other one I need guidance on is:

$\displaystyle\int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta$

$$\Rightarrow\int_{-\pi}^{\pi}\left(\left(\frac{1}{\sqrt{2}}\right)^2-\frac{\cos(2\theta)}{2}\right)^2\cos(2\theta) \ d\theta$$

$$\Rightarrow\int_{-\pi}^{\pi}\frac{\cos(4\theta)\cos(2\theta)}{4} \ d\theta$$

Now I am stuck again.

$$\int_{-\pi}^{\pi}\sin^4\theta\cos(2\theta) \ d\theta=$$
$$\int_{-\pi}^{\pi}\left[\frac{\cos(2\theta)}{4}-\frac{\cos^2(2\theta)}{2}+\frac{\cos(2\theta)* \cos^2(2\theta)}{4}\right] \ d\theta=$$
$$\int_{-\pi}^{\pi}\left[\frac{\cos(2\theta)}{4}-\left(\frac{1}{4}+\frac{\cos(4\theta)}{4}\right)+\frac{\cos(2\theta)}{8}+\frac{\cos(2\theta)*\cos(4\theta)}{8}\right] \ d\theta$$
$$\int_{-\pi}^{\pi}\left[\frac{3\cos(2\theta)}{8}-\frac{1}{4}-\frac{\cos(4\theta)}{4}+\frac{\cos(2\theta)*\cos(4\theta)}{8}\right] \ d\theta$$