# Integral of sin*sin

1. Apr 13, 2014

### Mary

Hey, I'm having trouble with one of the examples in my quantum book. I'm suppose to be showing that two eigenfunctions are orthogonal and in order to do that I have to solve the integral I have attached to this forum. I have the solution but I don't understand the steps! I believe it may be a trig substitution that I just can't remember. I would really appreciate any help. Thank you :)

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2. Apr 13, 2014

### lurflurf

$$\sin^2(x)=\left. \left. \dfrac{1}{2}\right( 1-\cos(x) \right) \\ \int_0^a \! \sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n \, \pi \, x}{a}\right)\mathrm{d}x=\dfrac{1}{a}\int_0^a \! \left( 1-\cos\left(\frac{2 \, n \, \pi \, x}{a} \right) \right) \mathrm{d}x$$

3. Apr 13, 2014

### Simon Bridge

Puzzled...
Is $\psi_n=\sqrt{\frac{2}{a}}\sin (n\pi x/a)$ orthogonal to $\psi_n$ ?
Or are you supposed to show that $\psi_n$ is orthogonal to $\psi_m$ where $m\neq n$?

Anyway - I'd add that you will do well to arm yourself with a table of trig identities:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities
... after a while you'll just remember the ones you use all the time.

4. Apr 13, 2014

### Mary

Ah, yes, I accidently put n$\pi$x /a for both eigen functions. One should be l$\pi$x /a. Thank you.

5. Apr 13, 2014

### Mary

Thank you for your quick response. I just realized that I accidently put n for both eigenfunctions. One of the sin's should be l$\pi$x/a. So, should the problem be different then?

6. Apr 13, 2014

### Mary

Ah, nevermind, I see now. After reviewing the trigometric identities on wikipedia I see how simple it is now. Thanks a bunch for all of your all's help!

7. Apr 14, 2014

### Simon Bridge

No worries, it's all good stuff :)