orthogonal chebyshev polyhomials This Theorem says
statement:
 $\int_{-1}^{1} \frac{T_n(x)T_m(x)}{ \sqrt{1+x^2} } dx = 0 ;$ when $n\ne m $
proof:
"substitute $x= cos \theta$ " 
and that's it.
So I am wondering should I start with this 
$\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+x^2}} dx $
or with this
$\int_{-1}^{1} \frac{cos(\theta n)cos(\theta m)}{\sqrt{1+cos^2 \theta}} dx $
I order to verify the proof.
note: T(x) is Chebyshev polynomial. 

When you do this, you apply the substitution rule for integrals, fully. Everything with $x$ in it, including the denominator, the $dx$, and the limits, transform.
On the other hand, the statement that you're trying to prove is incorrect. The correct form has $\sqrt{1-x^2}$ in the denominator instead.
