= Solution
{c}
Set $x=\cos\theta$ with $0\leq\theta\leq\pi$. The cosine addition identity gives
$$
\cos((n+1)\theta)+\cos((n-1)\theta)=2\cos\theta\cos(n\theta),
$$
hence $T_0=1$, $T_1=x$, and $T_{n+1}=2xT_n-T_{n-1}$. By <mathematical induction>, $T_n$ is a <polynomial> of degree $n$: the initial cases hold, and the leading term of $2xT_n$ has degree $n+1$ while the subtraction has lower degree. Its leading coefficient is $2^{n-1}$ for $n\geq1$.
The same change of variables gives the weighted <integral>
$$
\int_{-1}^1\frac{T_n(x)}{\sqrt{1-x^2}}\,dx=\int_0^\pi\cos(n\theta)\,d\theta
=\boxed{\begin{cases}\pi,&n=0,\\0,&n\geq1.\end{cases}}
$$
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