Legendre polynomials

Step 1: normalize $1$. $\langle 1,1\rangle=\int_{-1}^1 dt=2$, so
$$u_1=\frac{1}{\sqrt2}.$$

Step 2: orthogonalize $X$ against $u_1$. $\langle X,1\rangle=\int_{-1}^1 t\,dt=0$, so $X$ is already orthogonal to $1$. Norm:
$$\langle X,X\rangle=\int_{-1}^1 t^2\,dt=\frac23,\qquad u_2=\frac{X}{\sqrt{2/3}}=\sqrt{\frac32}\,X.$$

Step 3: orthogonalize $X^2$ against $u_1,u_2$.
$$\langle X^2,1\rangle=\int_{-1}^1 t^2\,dt=\frac23,\qquad \langle X^2,X\rangle=\int_{-1}^1 t^3\,dt=0.$$
So the orthogonal part is
$$X^2-\frac{\langle X^2,1\rangle}{\langle 1,1\rangle}\cdot 1=X^2-\frac{2/3}{2}=X^2-\frac13.$$
Its squared norm:
$$\int_{-1}^1\left(t^2-\frac13\right)^2 dt=\int_{-1}^1 t^4\,dt-\frac23\int_{-1}^1 t^2\,dt+\frac19\int_{-1}^1 dt=\frac25-\frac49+\frac29=\frac{18-20+10}{45}=\frac{8}{45}.$$
Hence
$$u_3=\frac{X^2-1/3}{\sqrt{8/45}}=\sqrt{\frac{45}{8}}\left(X^2-\frac13\right)=\frac{3\sqrt5}{2\sqrt2}\left(X^2-\frac13\right).$$

The orthonormal triple.
$$u_1=\frac{1}{\sqrt2},\qquad u_2=\sqrt{\frac32}\,X,\qquad u_3=\frac{3\sqrt5}{2\sqrt2}\left(X^2-\frac13\right).$$
They satisfy $\langle u_i,u_j\rangle=\delta_{ij}$.

Standard Legendre polynomials. Rescaling to monic (or to the classical normalization), one gets
$$P_0=1,\qquad P_1=X,\qquad P_2=\frac{3X^2-1}{2}.$$
These are proportional to $u_1,u_2,u_3$ and orthogonal for the integral inner product on $[-1,1]$.

Conclusion. The first three Legendre polynomials (unnormalized) are $P_0=1$, $P_1=X$, $P_2=\dfrac{3X^2-1}{2}$; the corresponding orthonormal vectors are $u_1,u_2,u_3$ above.