How can I change the variable in this integral?

Question

I have the following equality

$\int_{0}^{\pi}cos(nt)cos(mt)dt=0$ (if $m\neq n$)

This is in the context of Chebyshev polynomials and the book states the following to deduce ortoghonality.

"the orthogonality property is drawn by using the substitution t = arcos(x) in the interval [-1,1] relative to the weight function $(1-x^2)^{-\frac{1}{2}}$, which is verified by the system of polynomials {Tn(x)}:"

$\int_{-1}^{1}Tn(x)Tm(x)dt=0$, (if m $\neq$ n)

Can anybody explain (or point me to an example) how to do the above variable change in the initial integral and where does the weight function come from?

Answer 1

You can use the trigonometric product to sum formulae. $$ \cos A \cos B = \frac{1}{2}(\cos(A + B) + \cos(A - B))$$

Answer 2

$t = \arccos{x} \implies dt = -(1-x^2)^{-1/2} \, dx$.

Also note that

$$T_m(x) = \cos{(m \arccos{x})} $$

Answer 3

From $$\int_{0}^{\pi}\cos(nt)\cos(mt)dt=0,\qquad m\neq n$$ you may perform the change of variable $$ t= \text{arcos}(x), \qquad dt=-\frac{1}{\sqrt{1-x^2}}, $$ giving $$\int_{-1}^{1}T_n(x)T_m(x)dt=0,\qquad m\neq n.$$

Answer 4

It seems that you start from the result $\int_{\theta = 0}^{\theta = \pi}$ $cos(m \theta) cos(n \theta) d\theta = 0$ for $m \ne n$

Then if you use the suggested transformations and the property that $T_n(cos(\theta)) = cos(n \theta)$ you should be able to get

$\int_{x=-1}^{x=+1} T_m(x) T_n(x) \frac{1}{\sqrt{1-x^2}}dx= 0$ for $m \ne n$.

The factor $\frac{1}{\sqrt{1-x^2}}$ is the weight function.