Answer
$\sqrt{\frac{2}{L}}$
Work Step by Step
We need to find the normalization constant $ A_n $ for the wave functions of a particle in a rigid box which is given by
$$
\psi_n(x) = A \sin \left( \frac{n \pi x}{L} \right)
$$
The wave function must satisfy the normalization condition:
$$
\int_0^L |\psi_n(x)|^2 dx = 1
$$
This means that the probability of finding the particle anywhere in the box must be 1. So, we substitute the given form of $ \psi_n(x) $ into the normalization condition and solve for $ A $.
$$
\int_0^L \left[ A \sin \left( \frac{n \pi x}{L} \right) \right]^2 dx = 1
$$
We can take the constant $ A^2 $ outside of the integral:
$$
A^2 \int_0^L \sin^2 \left( \frac{n \pi x}{L} \right) dx = 1\tag 1
$$
Now we need to find the integral of
$$
\int_0^L \sin^2 \left( \frac{n \pi x}{L} \right) dx
$$
We can use the standard identity for the sine-squared function of
$\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$
So
$$
\int_0^L \sin^2 \left( \frac{n \pi x}{L} \right) dx = \frac{1}{2} \int_0^L \left[ 1 - \cos \left( \frac{2n \pi x}{L} \right) \right] dx
$$
$$
\int_0^L \sin^2 \left( \frac{n \pi x}{L} \right) dx=\frac{1}{2} \left[ \int_0^L 1 \; dx - \int_0^L \cos \left( \frac{2n \pi x}{L} \right) dx \right]\tag 2
$$
So the first integral is
$$
\int_0^L 1 \; dx = L\tag 3
$$
And 5he second integral is
$$
\int_0^L \cos \left( \frac{2n \pi x}{L} \right) dx = 0\tag 4
$$
Note that the integral of the cosine function over one complete period (from 0 to $ L $) is zero.
Plug (3) and (4) into (2),
$$
\int_0^L \sin^2 \left( \frac{n \pi x}{L} \right) dx=\frac{1}{2} \left[ L- 0\right]= \frac{L}{2}
$$
Plug into (1),
$$
A^2 \frac{L}{2} = 1
$$
Therefore,
$$
A = \sqrt{\frac{2}{L}}
$$
And hence
$$
\boxed{A_n = \sqrt{\frac{2}{L}}}
$$
