Deriving and Proving the Heat Equation's Integral Product

[squenshl]

Consider a heat equation for the temperature u of a rod of length 1:
u_t = u_xx, 0 < x < 1, t > 0 with boundary conditions u_x(0,t) = 0 & u(1,t) = 0. I derived X_n(x) = cos((n+1/2)\pix) using separation of variables.
How do I show that \int_{0}^1 X_n(x)X_m(x) dx = 1/2 if m = n and 0 if m \neq n.
I used the product to sum formula: cos(A)cos(B) = cos(A+B)/2 + cos(A-B)/2 to get 1/2cos((n+m+1)\pix) 1/2cos((n-m)\pix) but I am stuck after that. Someone help, am I even on the right track.

 

[LCKurtz]

Yes, you are on the right track. When m = n use the formula

\cos^2(\theta) = \frac {1 + \cos(2\theta)}{2}

which is easy to integrate. I'm not sure why you are stuck on the others.

\frac 1 2 \cos((n+m+1)\pi x)

is just as easy to integrate is \cos(kx).

n and m are integers, you know.