Example 1. Expand the function $y=\arccos x+1$ into a Fourier series on the interval $(-1,1)$ using the system of Chebyshev polynomials.
Solution. Chebyshev polynomials are defined by the formula

$$
T_{n}(x)=\cos (n \arccos x), \quad n=0,1,2, \ldots
$$

They form an orthogonal basis in the space of functions defined on the interval $(-1,1)$, with the scalar product

$$
(u, v)=\int_{-1}^{1} u(x) v(x) \frac{1}{\sqrt{1-x^{2}}} d x
$$

The desired expansion of the function $y=\arccos x+1$ in a Fourier series on the interval $(-1,1)$ with respect to the system $T_{0}(x), T_{1}(x), T_{2}(x), \ldots$ is given by

$$
\arccos x+1=\sum_{n=0}^{\infty} a_{n} T_{n}(x), \quad x \in(-1,1)
$$

where the coefficients $a_{0}, a_{1}, a_{2}, \ldots$ are determined by the Euler-Fourier formulas

$$
a_{n}=\frac{\left(\arccos x+1, T_{n}(x)\right)}{\left(T_{n}(x), T_{n}(x)\right)}=\frac{\int_{-1}^{1}(\arccos x+1) T_{n}(x) \frac{1}{\sqrt{1-x^{2}}} d x}{\int_{-1}^{1} T_{n}(x)^{2} \frac{1}{\sqrt{1-x^{2}}} d x}
$$

1. We check that

$$
\int_{-1}^{1}(\arccos x+1)^{2} d x<\infty
$$

The inequality holds because the function $y=\arccos x+1$ is bounded on $(-1,1)$.

2. We compute the coefficients $a_{0}, a_{1}, a_{2}, \ldots$ using the Euler-Fourier formulas (5).

The integrals

$$
\int_{-1}^{1}(\arccos x+1) T_{n}(x) \frac{1}{\sqrt{1-x^{2}}} d x
$$

are computed by making the substitution $x=\cos \vartheta$, then using formula (3) and integrating by parts. We obtain

$$
\begin{aligned}
\int_{-1}^{1}(\arccos x+1) T_{0}(x) \frac{1}{\sqrt{1-x^{2}}} d x & =\int_{0}^{\pi}(\vartheta+1) d \vartheta=\frac{\pi^{2}}{2}+\pi \\
\int_{-1}^{1}(\arccos x+1) T_{n}(x) \frac{1}{\sqrt{1-x^{2}}} d x & =\int_{0}^{\pi}(\vartheta+1) \cos n \vartheta d \vartheta=\frac{(-1)^{n}-1}{n^{2}}
\end{aligned}
$$

$n=1,2, \ldots$

We compute the integrals

$$
\int_{-1}^{1} T_{n}(x)^{2} \frac{1}{\sqrt{1-x^{2}}} d x=\int_{0}^{\pi} \cos ^{2} n \vartheta d \vartheta
$$

For $n=0$

$$
\int_{-1}^{1} T_{n}(x)^{2} \frac{1}{\sqrt{1-x^{2}}} d x=\pi
$$

For $n=1,2, \ldots$

$$
\int_{-1}^{1} T_{n}(x)^{2} \frac{1}{\sqrt{1-x^{2}}} d x=\frac{\pi}{2}
$$

Substituting the computed integrals into (5), we get

$$
a_{0}=\frac{\pi+2}{2}, \quad a_{n}=\frac{2}{\pi} \frac{(-1)^{n}-1}{n^{2}}, \quad n=1,2, \ldots
$$

3. Substituting the found values of the coefficients $a_{0}, a_{1}, a_{2}, \ldots$ into (4). We get

$$
\arccos x+1=\frac{\pi+2}{2}+\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2}} T_{n}(x)
$$

Answer. $\quad \arccos x+1=\frac{\pi+2}{2}+\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2}} T_{n}(x), x \in(-1,1)$