In Quantum Mechanics by Cohen-Tannoudji (volume 2, appendix 1), Cohen-Tannoudji gives a proof of the Plancherel theorem. I'm having trouble understanding this proof. Although this is a physics textbook, the question is mathematical and so I've used this site instead of the physics site. Cohen-Tannoudji defines the Fourier transform and its inverse for a 1D wave function as:
$$ \begin{align} \overline{\psi}(p) &= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx\, e^{-ipx/\hbar} \psi(x)\\ \psi(x) &= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dp\, e^{ipx/\hbar} \overline{\psi}(p) \end{align} $$
The Plancherel theorem states:
$$ \begin{equation} \int_{-\infty}^{\infty} dx |\psi(x)|^2 = \int_{-\infty}^{\infty} dp |\overline{\psi}(p)|^2 \end{equation} $$
Cohen-Tannoudji then gives the following proof ($z^*$ denotes a complex conjugate):
$$ \begin{align} \int_{-\infty}^{\infty} dx |\psi(x)|^2 &= \int_{-\infty}^{\infty} dx\, \psi^*(x) \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dp\, e^{ipx/\hbar} \overline{\psi}(p)\\ &= \int_{-\infty}^{\infty} dp\,\overline{\psi}(p) \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} dx\, e^{ipx/\hbar} \psi^*(x)\\ &= \int_{-\infty}^{\infty} dp\, \overline{\psi}^* (p) \overline{\psi}(p) \end{align} $$
The first line of the proof and the equality between the second and third lines are clear to me. However, I don't understand how the first and second lines are equal. My best guess for this is that he is somehow swapping $x$ and $p$. Why is line 2 equal to line 3?